Proof. Let G be the set of points $x\in M$ satisfying

$$ \begin{align*} \lim_{N\rightarrow\infty} \frac{S_N({\mathbf{1}}_{B_{2^{-l}}(x^*)})(x)}{N} = \mu(B_{2^{-l}}(x^*)) \quad \text{for all } l\in\mathbb{N}. \end{align*} $$

By the ergodic theorem, $\mu (G)=1$ . For $x\in G$ and $\varepsilon>0$ , choose l sufficiently large so that $2^{-l} < \varepsilon $ , then for all sufficiently large N, it follows that

$$ \begin{align*} S_N({\mathbf{1}}_{B_{2^{-l}}(x^*)})(x)> k(N). \end{align*} $$

This implies that all $k(N)$ closest points $\hat {x}_1,\ldots ,\hat {x}_{k(N)}$ lie in $B_{2^{-l}}(x^*)$ , so $d(\hat {x}_{k(N)},x^*) \leq 2^{-l}$ .

Proof. Applying the same argument as in Lemma 16 to the sets $\{f>R\}$ , where $R>0$ , instead of balls, yields that almost surely, $f(x_{k(N)})\rightarrow \infty $ as $N\rightarrow \infty $ . Keeping in mind that $f(y) \leq C_0 d(x^*,y)^{-\beta }$ for all points y, the claim follows.

Proof. By definition, $d(x^*,\hat {x}_{k(N)}) \leq d(x^*,x_{k(N)})$ , so it suffices to show

(16) $$ \begin{align} \limsup_{N\rightarrow\infty} \frac{d(x^*,x_{k(N)})}{d(x^*,\hat{x}_{k(N)})} \leq 1. \end{align} $$

If $d(x^*,\hat {x}_{k(N)})\ge d(x^*,x_{k(N)}),$ there is nothing to prove. Otherwise, $x_{k(N)}\notin \{\hat {x}_1,\ldots , \hat {x}_{k(N)}\}$ and, hence, there exists some $1\leq i\leq k(N)$ such that $\hat {x}_i\notin \{x_1,\ldots ,x_{k(N)}\}.$ By the definition of $x_1,\ldots ,x_{k(N)}$ , we have $f(\hat {x}_i)\leq f(x_{k(N)}).$ Moreover, $d(x^*,\hat {x}_i)\leq d(x^*,\hat {x}_{k(N)}).$ Thus, for all sufficiently large N,

$$ \begin{align*} \frac{\mathrm{Res}_{x^*}(f)(1-\epsilon(d(x^*,\hat{x}_{k(N)})))}{ d(x^*,\hat{x}_{k(N)})^\beta}&\leq f(\hat{x}_i) \leq f(x_{k(N)})\\ &\leq \frac{\mathrm{Res}_{x^*}(f)(1+\epsilon(d(x^*,x_{k(N)})))}{ d(x^*,x_{k(N)})^\beta }. \end{align*} $$

Since $d(x^*,\hat {x}_{k(N)})\to 0$ and $d(x^*,x_{k(N)})\to 0$ as $N\to \infty $ , we have

$$ \begin{align*} \epsilon(d(x^*,\hat{x}_{k(N)}))\to0 \quad\text{and}\quad \epsilon(d(x^*,x_{k(N)}))\to0, \end{align*} $$

and, hence, (16) follows.

Proof. Since $\hat {S}_N^{k(N)}(f) \geq S_N^{k(N)}(f)$ by definition, it suffices to show that

$$ \begin{align*} \limsup_{N\rightarrow \infty} \frac{\hat{S}_N^{k(N)}(f)(x)}{S_N^{k(N)}(f)(x)} \leq 1 \quad \text{for almost every } x. \end{align*} $$

Given that $d(x^*,\hat {x}_{k(N)}),\, d(x^*,x_{k(N)}) \rightarrow 0$ as $N\to \infty ,$ the desired inequality follows directly from (14).

Proof. By (17), $B_{r_N}(x^*)\subset A_N\subset B_{r_N+\delta _N}(x^*)$ and $\delta _N=o(r_N).$ Since $\mu (B_r(x^*))\sim B_d\rho (x^*)r^d$ and $r_N+\delta _N\sim r_N$ , the conclusion follows by squeezing.

Proof. Let $\widetilde r_N = r_N + \delta _N$ . By (17), $A_N \subset B_{\widetilde r_N}(x^*)$ and $\widetilde r_N \sim r_N$ . Hence, $|\!\log \widetilde r_N| \sim |\!\log r_N|$ and for large N, $1\le n \le C|\!\log r_N|$ implies $1\le n \le 2C|\!\log \widetilde r_N|$ . Applying the slow recurrence with exponent $A+2$ and time constant $2C$ yields

$$ \begin{align*} \mu(B_{\widetilde r_N}(x^*) \cap T^{-n}B_{\widetilde r_N}(x^*)) \le \mu(B_{\widetilde r_N}(x^*))\,|\!\log \widetilde r_N|^{-A-2}. \end{align*} $$

Since $\widetilde r_N \sim r_N$ and $\mu (B_{\widetilde r_N}(x^*)) \sim B_d\,\rho (x^*)\,\widetilde r_N^{\,d}$ , the right-hand side is bounded by $r_N^d|\!\log r_N|^{-A}$ for all sufficiently large N. The result follows from the inclusion $A_N \cap T^{-n}(A_N) \subset B_{\widetilde r_N}(x^*) \cap T^{-n}B_{\widetilde r_N}(x^*)$ .

Proof. We first record a local moment bound. Since $f(x) \le C_0\,d(x,x^*)^{- \beta }$ and $\rho $ is continuous, for any $s>0$ with $s\beta>d$ , there exists a constant $K_s$ independent of r, such that for all sufficiently small $r>0$ ,

(20) $$ \begin{align} \int_{M \setminus B_r(x^*)} f^s \, {d}\mu \le K_s \, r^{d - s\beta}. \end{align} $$

In addition, for small enough $r>0$ , it holds that

$$ \begin{align*} \mu (B_r(x^*)) \leq 2 \rho(x^*) B_d r^d. \end{align*} $$

Choose $q>1$ such that $q^2\beta>d$ and $\beta -d+({d}/{q^2})>0$ . Denote by p the conjugate exponent of q, and choose $H>0$ so large that $H(\beta -d+({d}/{q^2}))\ge A+2$ and $H(2\beta -d)\ge A+2$ .

Set $\hat {r}_N=r_N|\!\log r_N|^H$ and define $\hat {f}_N=f_N \cdot (1-{\mathbf {1}}_{B_{\hat {r}_N}(x^*)}).$ For $1\leq n <C |\!\log r_N|,$ we decompose the term $\mathbb {E}(f_N \cdot f_N\circ T^n)$ as

$$ \begin{align*} \mathbb{E}(f_N \cdot f_N\circ T^n)=\mathfrak{A} + \mathfrak{B} + \mathfrak{C} +\mathfrak{D}, \end{align*} $$

where

$$ \begin{align*} \begin{aligned} \mathfrak{A} &:= \int_{T^{-n}(B_{\hat{r}_N}(x^*))} \hat{f}_N \cdot f_N \circ T^n \, {d}\mu, \\ \mathfrak{B} &:= \int_{B_{\hat{r}_N}(x^*)} f_N \cdot \hat{f}_N \circ T^n \, {d}\mu, \\ \mathfrak{C} &:= \mathbb{E} (\hat{f}_N \cdot \hat{f}_N \circ T^n), \\ \mathfrak{D} &:= \int_{B_{\hat{r}_N}(x^*) \cap T^{-n}(B_{\hat{r}_N}(x^*))} f_N \cdot f_N \circ T^n\, {d}\mu. \end{aligned} \end{align*} $$

Bounding $\mathfrak {A}, \mathfrak {B}$ : applying Hölder’s inequality twice with exponents p and q yields

$$ \begin{align*} \mathfrak{A}&= \int_{T^{-n}(B_{\hat{r}_N}(x^*))} \hat{f}_N \cdot f_N \circ T^n \, {d}\mu\leq\mu(B_{\hat{r}_N}(x^*))^{{1}/{p}} (\mathbb{E}(\hat f_N^{q}\cdot f_N^{q}\circ T^n))^{1/q}\\ &\leq(2B_d\rho(x^*)\hat{r}_N^d)^{{1}/{p}} (\mathbb{E} (\hat{f}_N^{pq}))^{{1}/{(pq)}} (\mathbb{E}(f_N^{q^2}))^{1/q^2}. \end{align*} $$

Note that always $pq \geq 4$ ; hence, $p q \beta> d$ . Since $\hat {f}_N$ vanishes on $B_{\hat r_N}(x^*)$ and $f_N$ vanishes on $A_N\supset B_{r_N}(x^*)$ , we can apply (20) to obtain

$$ \begin{align*} \mathbb{E}(\hat{f}_N^{pq})\le K_{pq} \hat{r}_N^{d-pq\beta},\quad\mathbb{E}(f_N^{q^2})\le K_{q^2}r_N^{d-q^2\beta}. \end{align*} $$

For sufficiently large N, it follows that

$$ \begin{align*} \mathfrak{A}&\leq(2B_d\rho(x^*))^{{1}/{p}}K_{pq}^{{1}/{(pq)}} K_{q^2}^{{1}/{q^2}}\hat{r}_N^{-\beta+d-({d}/{q^2})} r_N^{-\beta+({d}/{q^2})}\\ &\leq(2B_d\rho(x^*))^{{1}/{p}}K_{pq}^{{1}/{(pq)}} K_{q^2}^{{1}/{q^2}}r_N^{-2\beta+d}|\!\log r_N|^{-A-2}. \end{align*} $$

The same estimate applies to $\mathfrak {B}.$

Bounding $\mathfrak {C}$ : we apply the Cauchy–Schwarz inequality and obtain

$$ \begin{align*} \mathfrak{C} \le \mathbb{E} (\hat{f}_N^2)\le K_2\hat{r}_N^{-2\beta+d}\le K_2 r_N^{-2\beta+d}|\!\log r_N|^{-A-2} \end{align*} $$

for N sufficiently large.

Bounding $\mathfrak {D}$ : let $A'>0$ be fixed, to be chosen later. By the slow recurrence property (see Definition 4), for N sufficiently large, for $1\le n<2C|\!\log \hat {r}_N|$ , it holds that

$$ \begin{align*} \mu(B_{\hat{r}_N}(x^*) \cap T^{-n}(B_{\hat{r}_N}(x^*))) \leq \mu(B_{\hat{r}_N}(x^*)) |\!\log \hat{r}_N|^{-A'} \leq 2B_d\rho(x^*)\hat{r}_N^d |\!\log \hat{r}_N|^{-A'}. \end{align*} $$

Applying Hölder’s inequality and (20), we obtain

$$ \begin{align*} \mathfrak{D} &\leq\mu(B_{\hat{r}_N}(x^*)\cap T^{-n}(B_{\hat{r}_N}(x^*)))^{{1}/{p}} (\mathbb{E} (f_N^q \cdot f_N^{q} \circ T^n ))^{{1}/{q}}\\ &\le(2B_d\rho(x^*)\hat{r}_N^d |\!\log \hat{r}_N|^{-A'})^{1/p}(\mathbb{E}(f_N^{pq}) )^{{1}/{(pq)}} (\mathbb{E} (f_N^{q^2}))^{{1}/{q^2}}\\ &\le(2B_d\rho(x^*))^{{1}/{p}}K_{pq}^{{1}/{(pq)}} K_{q^2}^{{1}/{q^2}}r_N^{-2\beta+d}|\!\log r_N|^{{Hd}/{p}}|\!\log\hat{r}_N|^{-({A'}/{p})}. \end{align*} $$

Note that $|\!\log \hat {r}_N|{\kern-1pt}\ge{\kern-1pt} \tfrac 12 |\!\log r_N|$ for sufficiently large N and choosing ${A'{\kern-1pt}>{\kern-1pt}Hd{\kern-1pt}+{\kern-1pt}p(A{\kern-1pt}+{\kern-1pt}2)}$ gives

$$ \begin{align*}\mathfrak{D}\leq 2^{{A'}/{p}}(2B_d\rho(x^*))^{{1}/{p}}K_{pq}^{{1}/{(pq)}} K_{q^2}^{{1}/{q^2}}r_N^{-2\beta+d}|\!\log r_N|^{-A-2}.\end{align*} $$

Combining the bounds for $\mathfrak {A}$ , $\mathfrak {B}$ , $\mathfrak {C}$ and $\mathfrak {D}$ , we obtain

$$ \begin{align*} \mathbb{E}(f_N \cdot f_N\circ T^n)\le\tilde{K}r_N^{-2\beta+d}|\!\log r_N|^{-A-2}, \end{align*} $$

where

$$ \begin{align*}\tilde{K}=(2+2^{{A'}/{p}})(2B_d\rho(x^*))^{{1}/{p}}K_{pq}^{{1}/{(pq)}} K_{q^2}^{{1}/{q^2}}+K_2. \end{align*} $$

Choose $N_0$ sufficiently large so that $\tilde {K}|\!\log r_N|^{-2}\le 1$ for all $N>N_0$ . Then,

$$ \begin{align*}\mathbb{E}(f_N \cdot f_N\circ T^n)\le r_N^{-2\beta+d}|\!\log r_N|^{-A}, \end{align*} $$

which completes the proof.

Proof of Proposition 25.

Denote $g_1={\mathbf {1}}_{A_N}$ and $g_2=f_N$ . By replacing f with $f/\mathrm {Res}_{x^*}(f)$ , we may assume throughout the proof that $\mathrm {Res}_{x^*}(f)=1$ . The general case is recovered by rescaling at the end.

Step 1: smooth approximation. Let $\epsilon _N>0$ be chosen later. Using (18), there is a $C^{\kappa }$ smooth function $h_1$ such that ${\mathbf {1}}_{A_N}\le h_1\le 1$ , with

$$ \begin{align*} \|g_1- h_1\|_{L^1} \ll \epsilon_N r_N^a \quad \text{and} \quad \|h_1\|_{C^{\kappa}} \ll \epsilon_N^{-\kappa} r_N^{-a'}. \end{align*} $$

Define $h_2= f \cdot (1-h_1)$ . Then,

$$ \begin{align*} \|g_2 - h_2 \|_{L^1} \ll \epsilon_N r_N^{a-\beta} \quad \text{and} \quad \|h_2\|_{C^{\kappa}} \ll \epsilon_N^{-\kappa} r_N^{-a'-\beta-\kappa}. \end{align*} $$

Define $\bar {g}_j=g_j-\int _M g_j \, {d}\mu $ and $\bar {h}_j=h_j - \int _M h_j \, {d}\mu $ for $j=1, 2.$ We next justify that replacing $g_j$ with $h_j$ in the integral of $\prod _{j=1}^2 (S_N(\bar {g}_j))^{m_j}$ introduces a negligible error.

Since centring preserves the $L^1$ -order, we have

(22) $$ \begin{align} \|\bar g_1-\bar h_1\|_{L^1}\ll \epsilon_N r_N^a \quad \text{and} \quad \|\bar g_2-\bar h_2\|_{L^1}\ll \epsilon_N r_N^{a-\beta}. \end{align} $$

We now estimate the error caused by replacing $g_j$ with $h_j$ . If $m_j=0$ , its error term below is omitted. For integers $r\ge1$ , using

$$ \begin{align*}|x^r-y^r|\le r|x-y|\max(|x|,|y|)^{r-1}, \end{align*} $$

we obtain

$$ \begin{align*} \begin{aligned} &\bigg\|\!\prod_{j=1}^2 (S_N(\bar g_j))^{m_j}-\prod_{j=1}^2 (S_N(\bar h_j))^{m_j}\bigg\|_{L^1}\\ &\quad\le m_1\|S_N(\bar g_1)-S_N(\bar h_1)\|_{L^1}\max(\|S_N(\bar g_1)\|_{L^\infty}, \|S_N(\bar h_1)\|_{L^\infty})^{m_1-1}\|S_N(\bar g_2)\|_{L^\infty}^{m_2} \\ &\qquad+m_2\|S_N(\bar g_2)-S_N(\bar h_2)\|_{L^1}\|S_N(\bar h_1)\|_{L^\infty}^{m_1} \max(\|S_N(\bar g_2)\|_{L^\infty},\|S_N(\bar h_2)\|_{L^\infty})^{m_2-1}. \end{aligned} \end{align*} $$

Using (22),

$$ \begin{align*}\|S_N(\bar g_1)-S_N(\bar h_1)\|_{L^1}\ll N\epsilon_N r_N^a,\quad \|S_N(\bar g_2)-S_N(\bar h_2)\|_{L^1}\ll N\epsilon_N r_N^{a-\beta}.\end{align*} $$

Moreover,

$$ \begin{align*}\|S_N(\bar g_1)\|_{L^\infty},\|S_N(\bar h_1)\|_{L^\infty}\ll N \quad\text{and}\quad \|S_N(\bar g_2)\|_{L^\infty},\|S_N(\bar h_2)\|_{L^\infty}\ll Nr_N^{-\beta}.\end{align*} $$

Therefore,

$$ \begin{align*}\bigg\|\!\prod_{j=1}^2 (S_N(\bar g_j))^{m_j}-\prod_{j=1}^2 (S_N(\bar h_j))^{m_j}\bigg\|_{L^1}\ll \epsilon_N N^m r_N^{a-m_2\beta}. \end{align*} $$

After division by the normalization in (21), this is bounded by $O(\epsilon _N N^{{m}/{2}}r_N^{a-({md}/{2})}).$ Since $Nr_N^d\to \infty $ , choosing $\epsilon _N=N^{-B}$ with $B>0$ sufficiently large makes this term $o(1)$ . Thus, the error caused by replacing $g_j$ by $h_j$ in (21) is negligible.

Therefore, the claim (21) will follow once we show

(23) $$ \begin{align} \lim_{N\rightarrow\infty} \frac{\int_M (S_N(\bar{h}_1))^{m_1} (S_N(\bar{h}_2))^{m_2}\, {d}\mu}{(B_d \rho(x^*))^{{m}/{2}} N^{{m}/{2}} r_N^{({md}/{2})-m_2\beta}}=\omega(m_1,m_2). \end{align} $$

Step 2: cluster decomposition. We expand the product

$$ \begin{align*} \prod_{j=1}^2 (S_N(\bar{h}_j))^{m_j} &= \sum_{\textbf{n}\in \{0,\ldots,N-1\}^m} \bigg(\prod_{i=1}^{m_1} (\bar{h}_1 \circ T^{n_i})\bigg) \bigg(\prod_{i=m_1+1}^m (\bar{h}_2 \circ T^{n_i})\bigg)\\ &=\sum_{\textbf{n}\in \{0,\ldots,N-1\}^m} \prod_{i=1}^{m} (\bar{h}_{j(i)} \circ T^{n_i}), \end{align*} $$

where

$$ \begin{align*} j(i)=\begin{cases} 1 & \text{if } 1\leq i \leq m_1,\\ 2 & \text{if } m_1 + 1 \leq i \leq m. \end{cases} \end{align*} $$

Let $ C> 0 $ denote a sufficiently large constant. For each $\textbf {n}= (n_1,\ldots , n_m)\in \{0,\ldots , N-1\}^m$ , define clusters as follows. There exists a minimal integer $1 \leq k=k(\textbf {n}) \leq m^2 + 1$ such that for all $i, i'\in \{1,\ldots ,m\},$

(24) $$ \begin{align} |n_i - n_{i'}| \leq C^k \log N \quad \text{or} \quad |n_i - n_{i'}|> C^{k+1} \log N. \end{align} $$

For $l\geq 1$ , we call $\{i_1,\ldots ,i_l\}\subset \{1,\ldots ,m\}$ a cluster of $\textbf {n}$ if it is a maximal collection with

$$ \begin{align*} |n_i - n_{i'}| \leq C^k \log N \quad \text{for all } i,i'\in \{i_1,\ldots,i_l\}. \end{align*} $$

In such cases, define $ l $ as the cluster length. Then, there exists an integer ${1{\kern-1pt}\leq{\kern-1pt} \eta {\kern-1pt}={\kern-1pt} \eta (\textbf {n}) {\kern-1pt}\leq{\kern-1pt} m}$ such that $ \textbf {n}$ consists of the clusters $\{i^s_1,\ldots , i^s_{l_s}\}$ for $ s = 1,\ldots , \eta $ , with corresponding lengths $ l_s $ .

Equivalently, let $ k $ be the smallest integer for which the set $\{n_i\}$ can be partitioned into $\eta $ clusters of sizes $l_s$ , each being a maximal subset where pairwise differences are at most $C^k \log N$ and distinct clusters are separated by more than $C^{k+1} \log N$ .

For each cluster, let $b_s=\min_{1\le r\le l_s}n_{i_r^s}$ . By exponential mixing, we have

(25) $$ \begin{align} &\bigg|\!\int \bigg(\prod_{i=1}^m (\bar{h}_{j(i)} \circ T^{n_i})\bigg) \, {d}\mu -\prod_{s=1}^{\eta}\int\bigg(\prod_{r=1}^{l_s} (\bar{h}_{j(i^s_r)} \circ T^{n_{i^s_r}})\bigg) \, {d}\mu\bigg|\nonumber\\ &\quad\ \ \ll e^{-\gamma C^{k+1} \log N}\prod_{s=1}^\eta\bigg\|\!\prod_{r=1}^{l_s} (\bar{h}_{j(i^s_r)} \circ T^{n_{i^s_r}-b_s})\bigg\|_{C^{\kappa}}. \end{align} $$

Each $\bar {h}_j$ satisfies $\|\bar {h}_j\|_{C^{\kappa }} \ll N^{B'}$ for a large constant $B'>0$ . Since the time differences within each cluster are at most $C^k \log N$ , we obtain

$$ \begin{align*} \bigg\|\!\prod_{r=1}^{l_s} (\bar{h}_{j(i^s_r)} \circ T^{n_{i^s_r}-b_s})\bigg\|_{C^{\kappa}} &\ll\prod_{r=1}^{l_s} \| \bar{h}_{j(i^s_r)} \|_{C^{\kappa}} L^{C^k \log N}\!\ll(N^{B'} L^{C^k \log N})^{l_s}, \end{align*} $$

where $L=\sup _{\|f\|_{C^{\kappa }}=1} \|f\circ T\|_{C^{\kappa }}$ is the operator norm of $f\mapsto f\circ T$ on $C^{\kappa }$ .

Consequently, the right-hand side of (25) can be estimated as

(26) $$ \begin{align} e^{-\gamma C^{k+1}\log N}N^{B'm}L^{mC^k\log N}\!\le N^{-100m}, \end{align} $$

which holds for sufficiently large C independent of $N.$ After summing over all $\mathbf n$ , this error is negligible.

Considering (23) and (26), the claim (21) will follow once we show

$$ \begin{align*} \lim_{N\rightarrow \infty} \frac{\sum_{\textbf{n}\in \{0,\ldots,N-1\}^m} \prod_{s=1}^{\eta(\textbf{n})} \int_M \prod_{r=1}^{l_s} (\bar{h}_{j(i^s_r)} \circ T^{n_{i^s_r}}) \, {d}\mu}{(B_d \rho(x^*))^{{m}/{2}} N^{{m}/{2}} r_N^{({md}/{2})-m_2\beta}} = \omega(m_1,m_2). \end{align*} $$

Replacing once more the smooth approximations $\bar {h}_{j(i^s_r)}$ by $\bar {g}_{j(i^s_r)}$ on the left-hand side, similarly to (23), only introduces another negligible error term of order $o(1)$ . Therefore, the proof will be finished once we show

(27) $$ \begin{align} \lim_{N\rightarrow \infty} \frac{\sum_{\textbf{n}\in \{0,\ldots,N-1\}^m} \prod_{s=1}^{\eta(\textbf{n})} \int_M \prod_{r=1}^{l_s} (\bar{g}_{j(i^s_r)} \circ T^{n_{i^s_r}})\, {d}\mu}{(B_d \rho(x^*))^{{m}/{2}} N^{{m}/{2}} r_N^{({md}/{2})-m_2\beta }} = \omega(m_1,m_2). \end{align} $$

Step 3: negligible cluster configurations. Define $\Delta _\eta = \{\mathbf {n} : \mathbf {n} \text { consists of } \eta \text { clusters}\}.$ It holds that

(28) $$ \begin{align} \#\Delta_{\eta} \ll \binom{m}{\eta}\eta !N^{\eta}\eta^{m-\eta} (2C^k\log N)^{m-\eta}\ll N^{\eta} (\log N)^{m}. \end{align} $$

Case 3.1: $\eta>({m}/{2}).$ At least one cluster is a singleton. Since

$$ \begin{align*}\int\bar{g}_{j(i)} \circ T^{n_i} \, {d}\mu = 0,\end{align*} $$

then it holds that

$$ \begin{align*} \prod_{s=1}^{\eta} \int_M \prod_{r=1}^{l_s} (\bar{g}_{j(i^s_r)} \circ T^{n_{i^s_r}}) \, {d}\mu=0. \end{align*} $$

Case 3.2: $\eta <({m}/{2}).$ If at least one cluster is a singleton, it can be treated as in Case 3.1. If no cluster is a singleton, then all $l_s \geq 2$ . Given that

$$ \begin{align*} \|\bar{g}_1\|_{L^2} \ll r_N^{{d}/{2}},\quad \|\bar{g}_1\|_{L^{\infty}} \ll 1, \quad \text{and}\\ \|\bar{g}_2\|_{L^2} \ll r_N^{-\beta+({d}/{2})},\quad \|\bar{g}_2\|_{L^{\infty}} \ll r_N^{-\beta}, \end{align*} $$

we have

(29) $$ \begin{align} \bigg|\!\prod_{s=1}^{\eta}\int\prod_{r=1}^{l_s}(\bar{g}_{j(i^s_r)} \circ T^{n_{i^s_r}})\, {d}\mu\bigg|\ll r_N^{-m_2 \beta + \eta d}. \end{align} $$

Thus,

$$ \begin{align*} \begin{aligned} \sum_{\eta<m/2} \sum_{\mathbf{n} \in \Delta_\eta} \bigg|\!\prod_{s=1}^{\eta}\int\prod_{r=1}^{l_s}(\bar{g}_{j(i^s_r)} \circ T^{n_{i^s_r}})\, {d}\mu\bigg| &\ll\sum_{\eta<m/2}N^{\eta} (\log N)^m r_N^{-m_2 \beta + \eta d}\\ &\ll\sum_{\eta<m/2} (N r_N^d)^\eta r_N^{- m_2 \beta}(\log N)^m. \end{aligned} \end{align*} $$

Subcase 3.2(i): $N r_N^d>N^{1/100}$ . This bound is of order $o(N^{{m}/{2}} r_N^{{md}/{2} - m_2 \beta }).$

Subcase 3.2(ii): $N r_N^d\leq N^{1/100}$ . For each index $\textbf {n}\in \Delta _{\eta }$ , we distinguish two cases.

Subcase 3.2(ii-a): all $n_i$ within the same cluster are equal for each cluster. Then, there exist $O(N^\eta )$ such indices n. Taking into account the estimate in (29), the total contribution from these indices is of order $O(N^{\eta } r_N^{\eta d - m_2 \beta }).$

Subcase 3.2(ii-b): at least one cluster contains two distinct indices. Then, Lemma 23 provides an additional saving. Using the estimates established in Case 4.1(B) below, (29) can be replaced by

$$ \begin{align*} \bigg| \prod_{s=1}^{\eta} \int \prod_{r=1}^{l_s} (\bar{g}_{j(i^s_r)} \circ T^{n_{i^s_r}} )\, {d}\mu \bigg| &\ll r_N^{\eta d - m_2 \beta} |\!\log r_N|^{-100m} \\ &\ll r_N^{\eta d - m_2 \beta} (\log N)^{-100m}. \end{align*} $$

Note that here, the assumption $N r_N^d\leq N^{{1}/{100}}$ is used crucially to show $\log N \asymp |\!\log r_N|$ .

Since the number of indices $\mathbf {n}$ in this subcase is at most $\#\Delta _{\eta }\ll N^{\eta } (\log N)^m$ , the total contribution from this case is of order

$$ \begin{align*} O(N^{\eta} r_N^{\eta d - m_2 \beta} (\log N)^{-99m}). \end{align*} $$

Combining Subcases 3.2(ii-a) and 3.2(ii-b), it holds that

$$ \begin{align*} \sum_{\eta < {m}/{2}} \sum_{\textbf{n} \in \Delta_{\eta}} \bigg| \prod_{s=1}^{\eta} \int \prod_{r=1}^{l_s} (\bar{g}_{j(i^s_r)} \circ T^{n_{i^s_r}}) \, {d}\mu\bigg| &\ll \sum_{\eta < {m}/{2}} N^{\eta} r_N^{\eta d - m_2 \beta}. \end{align*} $$

Using the assumption $N r_N^d \rightarrow \infty $ , it follows that $N^{\eta } r_N^{\eta d - m_2 \beta } =o( N^{{m}/{2}}\, r_N^{{md}/{2} - m_2 \beta })$ for each $\eta <({m}/{2})$ . Therefore, the above bound is of order $o( N^{{m}/{2}}\,r_N^{{md}/{2} - m_2 \beta })$ .

In conclusion, all terms in (27) for $\eta \neq {m}/{2}$ are asymptotically negligible. Note that, in particular, if m is odd, then (27) is satisfied. So, for the rest of the proof, assume m is even.

Step 4: the critical case $\eta ={m}/{2}$ . Unless all the clusters are of length $2$ , $\textbf {n}$ must have a singleton. Define

$$ \begin{align*} \Gamma=\{\mathbf{n}\in \Delta_{{m}/{2}} : \text{each cluster has size } 2\}. \end{align*} $$

Then, we consider $\sum _{\mathbf {n}\in \Gamma }\prod _{s=1}^{{m}/{2}}\int (\bar {g}_{j(i^s_1)} \cdot \bar {g}_{j(i^s_2)} \circ T^{|n_{i^s_2}-n_{i^s_1}|})\, {d}\mu .$

Since all terms with singletons are negligible—as in Case 3.1—it is enough to show

(30) $$ \begin{align} \lim_{N\rightarrow \infty} \frac{\sum_{\textbf{n}\in \Gamma} \prod_{s=1}^{{m}/{2}} \int_M \prod_{r=1}^{2} (\bar{g}_{j(i^s_r)} \circ T^{n_{i^s_r}}) \, {d}\mu}{(B_d \rho(x^*))^{{m}/{2}}\, N^{{m}/{2}}\, r_N^{({md}/{2})-m_2\beta }} = \omega(m_1,m_2). \end{align} $$

Step 4.1: estimates for a single pair. For $n_i,n_{i'}$ in the same cluster, we aim to estimate $\int _M (\bar {g}_{j(i)} \circ T^{n_i})(\bar {g}_{j(i')} \circ T^{n_{i'}})\, {d}\mu $ . Simplifying notation, we consider

$$ \begin{align*} \int_M \bar{g}_j \cdot \bar{g}_{j'} \circ T^k\, {d}\mu \end{align*} $$

for $j,j' \in \{1,2\}$ and $k=|n_i - n_{i'}| \leq C^{m^2+1} \log N,$ see (24). In the following, we will distinguish several cases.

We first highlight a technical point. Since $g_1 = {\mathbf {1}}_{A_N}$ , the slow recurrence estimate, specifically Lemma 23, is applicable only up to times $k=C' |\!\log r_N|$ for a constant $C'>0$ , which might be small if $r_N$ decays substantially slower than $1/N$ .

Case 4.1(A): $k=0$ . We consider the cases where $j=j'=1$ , $j=j'=2$ and $j\neq j'$ . For $j=j'=1$ , we use Lemma 22 to obtain

(31) $$ \begin{align} \int \bar{g}_1^2\, {d}\mu\sim \mu(A_N)\sim B_d \rho(x^*) r_N^d. \end{align} $$

For $j=j'=2$ , a straightforward calculation in Euclidean space reveals

(32) $$ \begin{align} \int_M \bar{g}_2^2 \, {d}\mu &\sim\int_{r_N \leq d(x,x^*) < 1} f^2(x)\, {d}\mu(x)\nonumber\\ &\sim \rho(x^*)\int_{\textbf{y}\in \mathbb{R}^d, r_N < \|\textbf{y}\| < 1} \frac{1}{\|\textbf{y}\|^{2\beta}} \, {d} \textbf{y}\nonumber\\ &\sim B_d \rho(x^*) r_N^{-2\beta+d} \int_1^{\infty} u^{-2\alpha} \, {d} u \nonumber\\ &\sim \sigma^2 B_d \rho(x^*) r_N^{-2\beta+d}, \end{align} $$

where $\sigma^2 = 1/(2 \alpha - 1)$ .

A straightforward computation in local coordinates near $x^*$ , together with the boundedness of f away from $x^*$ , gives

$$ \begin{align*} \|g_2\|_{L^1}\ll \begin{cases} 1 & \text{if } \beta<d,\\ |\!\log r_N| & \text{if } \beta=d,\\ r_N^{d-\beta} & \text{if } \beta>d. \end{cases} \end{align*} $$

For $j\neq j'$ , since the supports of $g_1$ and $g_2$ are disjoint, we therefore have

$$ \begin{align*} \bigg|\!\int_M \bar g_1\bar g_2\, {d}\mu\bigg|=\bigg|\!\int_M g_1\, {d}\mu\bigg|\bigg|\!\int_M g_2\, {d}\mu\bigg| &\ll r_N^{d}(1+|\!\log r_N|+r_N^{d-\beta})\\ &=r_N^{d-\beta} (r_N^{\beta}(1+|\!\log r_N|) + r_N^d). \end{align*} $$

Case 4.1(B): $0<k<C' |\!\log r_N|,$ where $C'>0$ is a sufficiently large constant. For $j=j'=1$ , applying Lemma 23 to $A_N \cap T^{-k}(A_N)$ with $A=100m$ yields

$$ \begin{align*} \bigg|\!\int_M \bar{g}_1 \cdot \bar{g}_1 \circ T^{k} \, {d}\mu\bigg|\ll \mu(A_N\cap T^{-k}A_N) + \mu(A_N)^2 \ll r_N^d |\!\log r_N|^{-100m}. \end{align*} $$

For $j=j'=2$ , Lemma 24 with $A=100m$ yields

(33) $$ \begin{align} \bigg|\!\int_M \bar{g}_2 \cdot \bar{g}_2 \circ T^k \, {d}\mu \bigg| \ll r_N^{-2\beta+d} |\!\log r_N|^{-100m}. \end{align} $$

For the cross term $j\neq j'$ , we consider the case $j=1$ and $j'=2$ ; the reverse situation follows analogously.

Let $\hat {r}_N=r_N|\!\log r_N|^H$ , where $H>({100m}/{\beta })$ is fixed. Since $A_N$ is regularly adapted, for sufficiently large N, we have $A_N\subset B_{\hat {r}_N}(x^*)$ . Hence,

$$ \begin{align*} \int_M g_1\cdot g_2\circ T^k\, {d}\mu = \int_{A_N\cap T^{-k}(M\setminus A_N)}f\circ T^k\, {d}\mu = \mathfrak A+\mathfrak B, \end{align*} $$

where

$$ \begin{align*} \mathfrak A&:= \int_{A_N\cap T^{-k}(B_{\hat{r}_N}(x^*)\setminus A_N)} f\circ T^k\, {d}\mu,\\ \mathfrak B&:= \int_{A_N\cap T^{-k}(M\setminus B_{\hat{r}_N}(x^*))} f\circ T^k\, {d}\mu. \end{align*} $$

For $\mathfrak A$ , note that for sufficiently large N, $|\!\log r_N|\leq 2|\!\log \hat {r}_N|$ . Since $k<C'|\!\log r_N|$ , we therefore have $k<2C'|\!\log \hat r_N|.$ Using slow recurrence with exponent $A'=Hd+100m$ and the inclusion $B_{r_N}(x^*)\subset A_N\subset B_{\hat r_N}(x^*),$ we obtain

$$ \begin{align*} \mathfrak A &\leq \sup_{y\notin B_{r_N}(x^*)}|f(y)| \mu(B_{\hat{r}_N}(x^*)\cap T^{-k}B_{\hat{r}_N}(x^*))\\ &\ll r_N^{-\beta}\hat{r}_N^d|\!\log\hat{r}_N|^{-A'}\\ &\ll r_N^{d-\beta}|\!\log r_N|^{-100m}. \end{align*} $$

For $\mathfrak B$ , using Lemma 22,

$$ \begin{align*}\mathfrak B\leq\sup_{y\notin B_{\hat{r}_N}(x^*)}|f(y)|\,\mu(A_N)\ll \hat{r}_N^{-\beta}r_N^d\ll r_N^{d-\beta}|\!\log r_N|^{-100m}. \end{align*} $$

Combining the estimates for $\mathfrak A$ and $\mathfrak B$ gives

$$ \begin{align*} \int_M g_1\cdot g_2\circ T^k\, {d}\mu \ll r_N^{d-\beta}|\!\log r_N|^{-100m}. \end{align*} $$

Finally, centring does not affect the estimate, so for $j\neq j'$ ,

$$ \begin{align*} \bigg|\!\int_M\bar g_j\cdot\bar g_{j'}\circ T^k\, {d}\mu\bigg| \ll r_N^{d-\beta}|\!\log r_N|^{-100m}. \end{align*} $$

Case 4.1(C): $k> C' |\!\log r_N|.$ We emphasize that this case only needs to be considered when $|\!\log r_N|=o(\log N)$ . Indeed, all computations here are still inside a cluster, so the relevant time differences satisfy $k=O(\log N)$ . If $|\!\log r_N|$ is comparable to $\log N$ , then taking $C'$ large enough makes the condition $k>C'|\!\log r_N|$ incompatible with the intra-cluster bound $k=O(\log N)$ and this subcase becomes empty. Using (18) with some $\varepsilon \in (0,r_N)$ to be determined later, there exists a $C^{\kappa }$ smooth function $\theta :M\rightarrow [0,1]$ such that

$$ \begin{align*} \|g_1-\theta\|_{L^1} \ll\varepsilon r_N^a \quad \text{and} \quad \|\theta\|_{C^{\kappa}} \ll\varepsilon^{-\kappa} r_N^{-a'}. \end{align*} $$

Denote $\bar {\theta }=\theta - \int _M \theta \, {d}\mu $ . By applying exponential mixing, we obtain

$$ \begin{align*} \bigg|\!\int_M \bar{g}_{1} \cdot \bar{g}_{1} \circ T^{k} \, {d}\mu \bigg| &\ll \bigg|\!\int_M \bar{\theta} \cdot \bar{\theta} \circ T^{k} \, {d}\mu \bigg| + \varepsilon r_N^{a} \\ & \ll e^{-\gamma k} \varepsilon^{-2\kappa} r_N^{-2a'} +\varepsilon r_N^a. \end{align*} $$

Choosing $\varepsilon = e^{-\gamma k/(2 \kappa+1)}$ guarantees $\varepsilon <r_N$ whenever $C'>(2\kappa +1)/\gamma $ , since ${k>C'|\!\log r_N|}$ . We then obtain

$$ \begin{align*} \bigg|\!\int_M \bar{g}_{1} \cdot \bar{g}_{1} \circ T^{k} \, {d}\mu \bigg| \ll e^{-\gamma k/(2 \kappa+1)} r_N^{-2a'}. \end{align*} $$

If $k= \lceil C' |\log r_N| \rceil + i$ for $i\in \mathbb{N}\cup\{0\},$ then we have

$$ \begin{align*} \bigg|\!\int_M \bar{g}_{1} \cdot \bar{g}_{1} \circ T^{k} \, {d}\mu \bigg| \ll r_N^{({\gamma C'}/({2\kappa+1}))-2a'}e^{-({\gamma i}/({2\kappa+1}))}\ll r_N^{100d} e^{-({\gamma i}/({2\kappa+1}))} \end{align*} $$

for large $C'$ .

The same argument, with the corresponding smooth approximations of $g_2$ and of the cross terms, gives

$$ \begin{align*} \bigg|\!\int_M \bar{g}_{2} \cdot \bar{g}_{2} \circ T^{k} \, {d}\mu \bigg| \ll r_N^{100d} e^{-({\gamma i}/({2\kappa+1}))} \end{align*} $$

and

$$ \begin{align*} \bigg|\!\int_M \bar{g}_{j} \cdot \bar{g}_{j'} \circ T^{k} \, {d}\mu \bigg| \ll r_N^{100d} e^{-({\gamma i}/({2\kappa+1}))} \end{align*} $$

for $j\neq j'$ .

Step 4.2: Negligibility outside the diagonal pair configurations. Let

$$ \begin{align*} \Gamma'=\bigg\{\textbf{n} \in \Gamma \;|\; n_{i^s_1}=n_{i^s_2} \text{ and } j(i^s_1)=j(i^s_2)\text{ for all } s=1,\ldots,\frac{m}{2}\bigg\}, \end{align*} $$

then, using the estimates from $j\neq j'$ in Case 4.1(A), and Cases 4.1(B) and 4.1(C) above, we obtain

$$ \begin{align*} &\sum_{\textbf{n} \in \Gamma\setminus \Gamma'} \prod_{s=1}^{{m}/{2}} \bigg|\!\int_M \bar{g}_{j(i^s_1)} \cdot \bar{g}_{j(i^s_2)} \circ T^{n_{i^s_2}-n_{i^s_1}} \, {d}\mu\bigg|\\ & \quad \ll N^{{m}/{2}}\, r_N^{({md}/{2})-m_2 \beta}(r_N^\beta(1+|\!\log r_N|)+r_N^d) \\ &\qquad + N^{{m}/{2}} |\!\log r_N|^{{m}/{2}} r_N^{({md}/{2}) - m_2 \beta} |\!\log r_N|^{-100m}\\ & \quad \ll N^{{m}/{2}}\, r_N^{({md}/{2})-m_2 \beta} |\!\log r_N|^{-99m}. \end{align*} $$

Note that here we used the fact that all cases, where one of the $|n_{i^s_1}-n_{i^s_2}|>C'|\!\log r_N|$ , are negligible by summing the estimate for i from Case 4.1(C).

Step 4.3: The diagonal pair contribution. Using (31) and (32), for $\mathbf {n}\in \Gamma '$ , it holds that

$$ \begin{align*} \prod_{s=1}^{{m}/{2}} \int_M \bar{g}_{j(i^s_1)}^2 \, {d}\mu\sim(B_d \rho(x^*))^{{m}/{2}}\, r_N^{({md}/{2})-m_2\beta } \sigma^{m_2}. \end{align*} $$

If either $m_1$ or $m_2$ is odd, then $\Gamma '=\varnothing $ , since indices of type $g_1$ and $g_2$ must be paired separately. Otherwise, there are $(m_1-1)!!$ and $(m_2-1)!!$ possible pairings within each type. For each fixed pairing, this leaves $m/2$ free time variables; hence, $N^{{m}/{2}}$ choices. The cluster-separation condition excludes only configurations where two such times lie within distance $O(\log N)$ , whose number is $O(N^{({m}/{2})-1}\log N)$ . Hence,

$$ \begin{align*}\#\Gamma'\sim N^{{m}/{2}}(m_1-1)!!(m_2-1)!!. \end{align*} $$

It follows that

$$ \begin{align*} \sum_{\mathbf{n} \in \Gamma'} \prod_{s=1}^{{m}/{2}} \int_M \bar{g}_{j(i^s_1)}^2 \, {d}\mu \sim (B_d \rho(x^*))^{{m}/{2}} N^{{m}/{2}}\, r_N^{({md}/{2})-m_2\beta } \omega(m_1,m_2), \end{align*} $$

which completes the proof of (30).

Proof. We observe that

$$ \begin{align*}| \hat{S}_N^K(f) - S_N(f_N) | = \bigg|\!\sum_{n=0}^{N-1} f(T^n x) {\mathbf{1}}_{B_{r_N}(x^*)}(T^n x) - \sum_{i=1}^K f(\hat{x}_i) \bigg|.\end{align*} $$

If $S_N({\mathbf {1}}_{B_{r_N}(x^*)})(x)=m\leq K,$ then $\hat {x}_1,\ldots ,\hat {x}_m \in B_{r_N}(x^*)$ , where $\hat {x}_j$ is the jth closest orbit point to $x^*$ as defined in §3.1, and consequently,

$$ \begin{align*} | \hat{S}_N^K(f)(x) - S_N(f_N)(x) | \leq \sum_{i=m+1}^K f(\hat{x}_i) \leq \frac{KC_0N\log N}{(\log \log N)^6}. \end{align*} $$

It follows that

$$ \begin{align*}\bigg\{|\hat{S}_N^K(f) - S_N(f_N)|>\frac{KC_0N\log N}{(\log \log N)^6}\bigg\}\subset \{S_N({\mathbf{1}}_{B_{r_N}(x^*)}) \geq K+1\}.\end{align*} $$

Applying Markov’s inequality, we obtain for sufficiently large $N,$

(35) $$ \begin{align} &\mu\bigg(| \hat{S}_N^K(f) - S_N(f_N) |>\frac{KC_0N\log N}{(\log \log N)^6}\bigg)\nonumber\\ & \quad \leq\mu( S_N({\mathbf{1}}_{B_{r_N}(x^*)}) \geq K+1 ) \nonumber\\ & \quad \leq \mu(S_N({\mathbf{1}}_{B_{r_N}(x^*)})(S_N({\mathbf{1}}_{B_{r_N}(x^*)})-1)\geq (K+1)K)\nonumber\\ & \quad \leq \frac{2}{(K+1)K}\mathbb{E}\bigg(\binom{S_N({\mathbf{1}}_{B_{r_N}(x^*)})}{2}\bigg). \end{align} $$

We compute the second factorial moment

$$ \begin{align*} \mathbb{E}\bigg( \binom{S_N(\mathbf{1}_{B_{r_N}(x^*)})}{2} \bigg) = \sum_{0 \leq j < k \leq N-1} \mu( T^{-j} B_{r_N}(x^*) \cap T^{-k} B_{r_N}(x^*) ) \end{align*} $$

and consider two cases based on the distance between indices j and $k.$

Case 1 (long range): if $k-j> C \log N$ for some large constant $C>0$ , using Lemma 20 and the fact that $\mu (B_{r_N}(x^*)) \asymp (\log \log N)^6 / (N \log N)$ , we have

$$ \begin{align*} &\mu( T^{-j} B_{r_N}(x^*) \cap T^{-k} B_{r_N}(x^*))\\ &\quad = \mu(B_{r_N}(x^*))^2 + O( \mu(B_{r_N}(x^*))^{(({d-1})/{d}) \cdot ({2\kappa}/({2\kappa+1}))} N^{-({C \gamma}/({2\kappa+1}))} ). \end{align*} $$

Thus, the contribution from long-range pairs is $O( (\log \log N)^{12} / (N^2(\log N)^2) ).$ Since the number of such pairs is at most $N^2$ , their total contribution is

$$ \begin{align*}O\bigg(\frac{(\log \log N)^{12}}{(\log N)^2} \bigg).\end{align*} $$

Case 2 (short range): by the slow recurrence property with $A = 3$ , for large N, we obtain

$$ \begin{align*} \mu( T^{-j} B_{r_N}(x^*) \cap T^{-k} B_{r_N}(x^*)) \ll \mu(B_{r_N}(x^*)) (\log N)^{-2} \ll \frac{(\log\log N)^6}{N (\log N)^3}. \end{align*} $$

The number of such pairs is at most $C N \log N$ , so their total contribution is

$$ \begin{align*} O\bigg( \frac{(\log\log N)^6}{(\log N)^2} \bigg). \end{align*} $$

Combining both cases, we have

$$ \begin{align*} \mathbb{E}\bigg( \binom{S_N(\mathbf{1}_{B_{r_N}(x^*)})}{2} \bigg) \ll \frac{(\log \log N)^{12}}{(\log N)^2} + \frac{(\log\log N)^6}{(\log N)^2} \ll (\log N)^{-({3}/{2})}. \end{align*} $$

Finally, we complete the proof using the inequality (35).

Proof. First, note that

$$ \begin{align*}\mathbb{E} f_N \sim \operatorname{Res}_{x^*}(f)B_d \rho(x^*) \log N.\end{align*} $$

Let $\varepsilon _N \to 0$ be chosen such that

(36) $$ \begin{align} |\mathbb{E}f_N- \operatorname{Res}_{x^*}(f)B_d \rho(x^*) \log N|<\varepsilon_N\log N. \end{align} $$

Letting $\epsilon _N=2\varepsilon _N + (\log \log N)^{-1}$ , Chebyshev’s inequality yields

$$ \begin{align*} & \mu(| S_N(f_N)- \operatorname{Res}_{x^*}(f)B_d \rho(x^*) N \log N|> \epsilon_N N \log N)\\ & \quad \leq \mu \bigg(| S_N(f_N) - \mathbb{E} S_N(f_N) | > \frac{\epsilon_N N \log N}{2} \bigg) \ll \frac{\mathrm{Var}(S_N(f_N))}{(\epsilon_N N \log N)^2}. \end{align*} $$

Applying (20) with $s=2$ and $\beta =d$ , we obtain

$$ \begin{align*}\mathbb{E} f_N^2=\int_{M\setminus B_{r_N}(x^*)}f^2\, {d}\mu\ll r_N^{-d}. \end{align*} $$

For $1\le k\le C'|\!\log r_N|$ , Lemma 24 with $A=3$ gives

$$ \begin{align*}\mathbb{E}(\bar{f}_N\cdot\bar{f}_N\circ T^k)\le \mathbb{E}(f_N \cdot f_N\circ T^k)\ll r_N^{-d}|\!\log r_N|^{-3}, \end{align*} $$

where $\bar {f}_N=f_N-\mathbb {E} f_N.$

For $k=C'|\!\log r_N|+i$ , $i\ge 1$ , it follows from Case 4.1(C) in the proof of Proposition 25 that for $C'$ sufficiently large,

$$ \begin{align*}\mathbb{E}(\bar{f}_N\cdot\bar{f}_N\circ T^k)\ll r_N^{100d} e^{-({\gamma i}/({2\kappa+1}))}. \end{align*} $$

Consequently,

$$ \begin{align*} \operatorname{Var}(S_N(f_N))&=N\mathrm{Var}(f_N) +2\sum_{k=1}^{N-1}(N-k)\mathbb{E}(\bar{f}_N\cdot\bar{f}_N\circ T^k)\\ &\ll Nr_N^{-d}+N\sum_{1\le k\le C'|\!\log r_N|}r_N^{-d}|\!\log r_N|^{-3}+N\sum_{i\ge1} r_N^{100d}e^{-({\gamma i}/({2\kappa+1}))}\\ &\ll Nr_N^{-d}\ll\frac{N^2 \log N}{(\log \log N)^6} \end{align*} $$

and we conclude

$$ \begin{align*} \begin{aligned} & \mu(|S_N(f_N)- \mathrm{Res}_{x^*}(f)B_d \rho(x^*) N \log N|>\epsilon_N N \log N )\\ & \quad \ll \frac{N^2 \log N}{(\log \log N)^6(\epsilon_N N \log N)^2}\\ & \quad \ll (\log N)^{-1} (\log \log N)^{-4}. \end{aligned}\\[-37pt] \end{align*} $$

Proof of Theorem 5.

By Lemmas 27 and 28, for

$$ \begin{align*}\omega_N=\epsilon_N+\frac{KC_0}{(\log\log N)^6}\to 0,\end{align*} $$

we obtain

$$ \begin{align*} \mu\bigg(\bigg| \frac{\hat{S}_N^K(f)}{N \log N}-\operatorname{Res}_{x^*}(f)B_d \rho(x^*) \bigg|>\omega_N \bigg) \ll (\log N)^{-1} (\log \log N)^{-4}. \end{align*} $$

Now, consider the subsequence defined by

$$ \begin{align*}N_l = \bigg\lceil \exp\bigg(\frac{l}{(\log l)^2}\bigg) \bigg\rceil.\end{align*} $$

From the above, we obtain

$$ \begin{align*} \sum_{l=2}^\infty & \mu\bigg(\bigg|\frac{\hat{S}_{N_l}^K(f)}{N_l \log N_l}-\operatorname{Res}_{x^*}(f)B_d \rho(x^*) \bigg|>\omega_{N_l} \bigg)\\ & \ll \sum_{l=2}^\infty (\log N_l)^{-1}(\log\log N_l)^{-4}\\ &\ll\sum_{l=2}^{\infty} l^{-1}(\log l)^{-2}\\ &< \infty. \end{align*} $$

Therefore, by the first Borel–Cantelli lemma, for $\mu $ -almost every $x,$

$$ \begin{align*} \lim_{l\to \infty} \frac{\hat{S}_{N_l}^K(f)(x)}{N_l \log N_l} = \mathrm{Res}_{x^*}(f)B_d \rho(x^*). \end{align*} $$

For arbitrary $N \in [N_l, N_{l+1}],$ the monotonicity of the trimmed sum $\hat {S}_N^K(f)$ implies

$$ \begin{align*} \frac{\hat{S}_{N_l}^K(f)}{N_{l+1} \log N_{l+1}} \leq \frac{\hat{S}_N^K(f)}{N \log N} \leq \frac{\hat{S}_{N_{l+1}}^K(f)}{N_l \log N_l}. \end{align*} $$

Taking the limit as $l \to \infty $ and observing that $\lim _{l \to \infty }N_{l+1}/N_l = 1$ , we conclude that

$$ \begin{align*} \lim_{N \to \infty} \frac{\hat{S}_N^K(f)(x)}{N \log N} = \operatorname{Res}_{x^*}(f)B_d \rho(x^*) \end{align*} $$

for $\mu $ -almost every x. Finally, Lemma 19 implies that also

$$ \begin{align*} \lim_{N \to \infty} \frac{S_N^K(f)(x)}{N \log N} = \operatorname{Res}_{x^*}(f)B_d \rho(x^*).\\[-40pt] \end{align*} $$

Proof of Corollary 7.

Theorem 5 establishes that the normalized trimmed sum $S_N^K(f)/(N \log N)$ converges almost surely to $\operatorname {Res}_{x^*}(f)B_d \rho (x^*) $ . To pass from trimmed to untrimmed sums, we observe that the contribution from the K largest terms is negligible in probability:

$$ \begin{align*} &\mu\bigg(\sum_{i=1}^{K} f(x_i^N)> KN \log \log N \bigg)\leq \mu\bigg(\max_{0\le n<N}f(T^nx)>N\log\log N \bigg)\\ & \quad \leq\sum_{n=0}^{N-1}\mu(f\circ T^n>N\log\log N)= N\mu(f>N\log\log N)\ll (\log \log N)^{-1} \to 0. \end{align*} $$

Therefore, the difference between $S_N(f)/(N \log N)$ and $S_N^K(f)/(N \log N)$ vanishes in probability, and the full sum shares the same limit in probability, that is,

$$ \begin{align*} \lim_{N \to \infty} \mu\bigg(\bigg| \frac{S_N(f)}{N \log N}-\operatorname{Res}_{x^*}(f)B_d \rho(x^*)\bigg|> \varepsilon \bigg) = 0 \quad \text{ for all } \varepsilon > 0.\\[-41pt] \end{align*} $$

Proof. For even m, the conclusion follows from Proposition 25 applied with the regularly adapted sequence $A_N=B_{r_N}(x^*),$ and the choices $m_1=0$ , $m_2=m.$ If m is odd, set ${q=m+1}$ . Then,

$$ \begin{align*}\mathbb{E}(|S_N(\bar f_N)|^m)\le (\mathbb{E}(|S_N(\bar f_N)|^q))^{{m}/{q}}. \end{align*} $$

Applying the even case to q gives

$$ \begin{align*}\mathbb{E}(|S_N(\bar f_N)|^m)\ll(N^{{q}/{2}}\, r_N^{({qd}/{2})-q\beta})^{{m}/{q}}=N^{{m}/{2}}\, r_N^{({md}/{2})-m\beta},\end{align*} $$

which proves the lemma.

Proof. For even m, the result follows by applying Proposition 25 with the regularly adapted sequence $A_N=B_{r_N}(x^*)$ and parameters $m_1=m,m_2=0$ . The odd case then follows from Hölder’s inequality, exactly as in the proof of Lemma 29.

Proof. By the continuity of the density function of $\mu ,$ we have

$$ \begin{align*}r_N^d\sim \frac{k(N)}{B_d \rho(x^*) N}.\end{align*} $$

Define the event

$$ \begin{align*} E_N=\{|S_N({\mathbf{1}}_{B_{r_N}(x^*)})-k(N)|<k(N)^{1-\epsilon} \}\cap\{ S_N({\mathbf{1}}_{B_{r_N/2}(x^*)})<k(N)\}. \end{align*} $$

On $E_N$ , the difference between the trimmed sum and the truncated sum satisfies

$$ \begin{align*} |\hat{S}_N^{k(N)}(f)-S_N(f_N)|<C_02^{\beta} r_N^{-\beta} k(N)^{1-\epsilon}< C N^{\alpha} k(N)^{1-\alpha - \epsilon}, \end{align*} $$

where $C_0> 0$ is the constant from (11) and $C=C_0(2 B_d\rho (x^*)+10)^{\beta }.$ It follows that

$$ \begin{align*}\{|\hat{S}_N^{k(N)}(f)-S_N(f_N) |>C N^{\alpha} k(N)^{1-\alpha - \epsilon}\}\subset E_N^c.\end{align*} $$

The probability of the complement $E_N^c$ is bounded by

(37) $$ \begin{align} \mu(E_N^c)&\leq \mu(|S_N({\mathbf{1}}_{B_{r_N}(x^*)}) - k(N)| \geq k(N)^{1-\epsilon} )\nonumber\\ & \quad + \mu( |S_N({\mathbf{1}}_{B_{r_N/2}(x^*)}) - N \mu(B_{r_N/2}(x^*)) | \geq k(N) - N \mu(B_{r_N/2}(x^*)))\nonumber\\ &\leq \mu(|S_N({\mathbf{1}}_{B_{r_N}(x^*)})-k(N)| \geq k(N)^{1-\epsilon})\nonumber\\ & \quad + \mu\bigg(|S_N({\mathbf{1}}_{B_{r_N/2}(x^*)}) - N\mu(B_{r_N/2}(x^*))| \geq \frac{k(N)}{3} \bigg), \end{align} $$

where in the last step, we used the fact that $N \mu (B_{r_N/2}(x^*))\sim 2^{-d} k(N)$ .

By Lemma 30, for any integer $m \geq 2$ ,

$$ \begin{align*} \mathbb{E}(|S_N({\mathbf{1}}_{B_{r_N}(x^*)})-k(N)|^m)& \ll k(N)^{{m}/{2}} \\ \mathbb{E} (|S_N({\mathbf{1}}_{B_{r_N/2}(x^*)})-N\mu(B_{r_N/2}(x^*))|^m)& \ll k(N)^{{m}/{2}}. \end{align*} $$

Therefore, applying Markov’s inequality to (37), we deduce

$$ \begin{align*} \mu(E_N^c) \ll k(N)^{-({m}/{2})+m\epsilon} + k(N)^{-({m}/{2})} \ll k(N)^{-A}, \end{align*} $$

provided that $m>2A/(1-2\epsilon ).$ This completes the proof.

Proof. For $m\geq 2$ , Lemma 29 yields

$$ \begin{align*} \mathbb{E}(|S_N(\bar{f}_N)|^m) \ll N^{{m}/{2}}\, r_N^{- m\beta + ({md}/{2})} \ll N^{m\alpha} k(N)^{({m}/{2})-m\alpha}, \end{align*} $$

where $\bar {f}_N = f_N - \mathbb {E} f_N$ . Using Markov’s inequality, it follows that

$$ \begin{align*} \mu(| S_N(f_N) -\mathbb{E} S_N(f_N) |> N^{\alpha} k(N)^{1-\alpha-\epsilon}) & \ll \frac{\mathbb{E}(|S_N(\bar{f}_N)|^m)}{N^{m\alpha} k(N)^{m-m\alpha-m\epsilon}}\\ & \ll k(N)^{-({m}/{2}) + m \epsilon}. \end{align*} $$

The desired result follows by selecting an integer $m>2A/(1-2\epsilon ).$

Proof of Theorem 8.

Using Lemmas 31 and 32, for $A>0$ , we have

(38) $$ \begin{align} \mu (| \hat{S}_N^{k(N)}(f)-\mathbb{E} S_N(f_N) |>2C N^{\alpha} k(N)^{1-\alpha-\epsilon}) \ll k(N)^{-A}, \end{align} $$

where the constant $C>1$ is the one specified in Lemma 31.

Define the subsequence $\tilde {N}_l = \lceil 2^{l^{\epsilon }} \rceil $ and construct $N_{l,j}$ recursively by setting $N_{l,0} = \tilde {N}_l$ and $N_{l,j+1} = \min (\tilde {N}_{l+1},N^*),$ where

$$ \begin{align*} N^* = \max\bigg(N_{l,j} + 1, \;\max\bigg(N\geq N_{l,j}: \;\frac{k(N)}{k(N_{l,j})} < 2^{l^{-\epsilon}}\bigg)\bigg). \end{align*} $$

This ensures that the sequence $N_{l,j}$ increases monotonically from $\tilde {N}_l$ to $\tilde {N}_{l+1},$ and that the growth of $k(N)$ remains controlled at each step. Specifically, if $N_{l,j+2}<\tilde {N}_{l+1}$ , then

(39) $$ \begin{align} \frac{k(N_{l,j+2})}{k(N_{l,j})} \geq 2^{l^{-\epsilon}}. \end{align} $$

Furthermore, since $k(\tilde {N}_{l+1}) \leq \tilde {N}_{l+1} \leq 2^{(l+1)^{\epsilon }} + 1$ and the cumulative growth factor over $2 l^{3\epsilon }$ steps is at least $(2^{l^{-\epsilon }})^{l^{3\epsilon }} = 2^{l^{2\epsilon }},$ the sequence must reach $ \tilde {N}_{l+1}$ within at most $2 l^{3\epsilon }$ steps. Therefore,

$$ \begin{align*} N_{l,2 l^{3\epsilon}} = \tilde{N}_{l+1}. \end{align*} $$

Let $\{N_i\}$ be the ordered sequence of distinct values from $N_{l,j}$ . Then, for sufficiently small $\epsilon>0$ and $A>2 \epsilon ^{-2},$

(40) $$ \begin{align} \sum_{i=1}^{\infty} k(N_i)^{-A} \ll \sum_{l=1}^{\infty} l^{3\epsilon} (\log\tilde{N}_l)^{-A\epsilon} \ll \sum_{l=1}^{\infty} l^{-A \epsilon^2 + 3\epsilon} <\infty, \end{align} $$

because $\lim _{N\rightarrow \infty } {k(N)}/{(\log N)^{\epsilon }} = \infty $ . By the first Borel–Cantelli lemma and (38), we have

(41) $$ \begin{align} \lim_{i\rightarrow \infty} \frac{\hat{S}_{N_i}^{k(N_i)}(f)(x)}{\mathbb{E} S_{N_i}(f_{N_i})}=1 \quad \text{for almost every } x. \end{align} $$

Since

$$ \begin{align*} \mathbb{E} S_N(f_N) \sim c N^\alpha k(N)^{1-\alpha}, \end{align*} $$

with $c={1}/({\alpha -1}) \operatorname {Res}_{x^*}(f)B_d^{\alpha } \rho (x^*)^{\alpha }$ , it follows that

(42) $$ \begin{align} \lim_{i\rightarrow \infty} \frac{\hat{S}_{N_i}^{k(N_i)}(f)(x)}{N_i^{\alpha} k(N_i)^{1-\alpha}} = c \quad \text{for almost every } x. \end{align} $$

Define two auxiliary trimming sequences as

$$ \begin{align*} k_+(N) &= \lceil 2^{l^{-\epsilon}} k(N) \rceil \quad \text{if}\, \tilde{N}_l \leq N < \tilde{N}_{l+1}, \\ k_-(N) &= \lfloor 2^{-l^{-\epsilon}} k(N) \rfloor \quad \text{if}\, \tilde{N}_l \leq N < \tilde{N}_{l+1}. \end{align*} $$

Since $\lim _{N\rightarrow \infty } {k_{\pm }(N)}/{k(N)} = 1$ , Lemmas 31 and 32, as well as (40), also hold for $k_\pm $ instead of k; thus, the first Borel–Cantelli lemma implies

(43) $$ \begin{align} \lim_{i\rightarrow \infty} \frac{\hat{S}_{N_i}^{k_{\pm}(N_i)}(f)(x)}{N_i^{\alpha} k(N_i)^{1-\alpha}} = c \quad \text{for almost every } x. \end{align} $$

For $N \notin \{N_i\}_{i=1}^\infty $ with $\tilde {N}_l<N_i<N<N_{i+1}<\tilde {N}_{l+1},$ the construction of $\{N_i\}$ and $k_\pm $ implies

(44) $$ \begin{align} 2^{-l^{-\epsilon}} k_+(N_i)-1\leq k(N_i)\leq k(N)\leq k(N_{i+1})\leq 2^{l^{-\epsilon}} k(N_i) \leq k_+(N_i) \end{align} $$

and

(45) $$ \begin{align} k_-(N_{i+1}) \leq k(N) \leq 2^{l^{-\epsilon}} k_-(N_{i+1})+2^{l^{-\epsilon}}. \end{align} $$

Let x satisfy (42) and (43). By the monotonicity of trimmed sums and applying (44), (45), we obtain

$$ \begin{align*} \hat{S}_{N_i}^{k_+(N_i)}(f) \leq \hat{S}_N^{k(N)}(f) \leq \hat{S}_{N_{i+1}}^{k_-(N_{i+1})}(f).\end{align*} $$

Rewriting this inequality as

$$ \begin{align*} \frac{N_i^{\alpha}k_+(N_i)^{1-\alpha}}{N^{\alpha}k(N)^{1-\alpha}}\cdot \frac{\hat{S}_{N_i}^{k_+(N_i)}(f)}{N_i^{\alpha}k_+(N_i)^{1-\alpha}}&\leq \frac{\hat{S}_N^{k(N)}(f)}{N^{\alpha}k(N)^{1-\alpha}}\\ &\leq \frac{N_{i+1}^{\alpha}k_-(N_{i+1})^{1-\alpha}}{N^{\alpha}k(N)^{1-\alpha}}\cdot \frac{\hat{S}_{N_{i+1}}^{k_-(N_{i+1})}(f)}{N_{i+1}^{\alpha}k_-(N_{i+1})^{1-\alpha}} \end{align*} $$

and observing that the scaling factors converge to $1$ as $N \to \infty $ , it follows from the squeeze theorem that

$$ \begin{align*} \lim_{N\rightarrow\infty, N \not \in \{N_i\}} \frac{\hat{S}_N^{k(N)}(f)(x)}{N^{\alpha} k(N)^{1-\alpha}}=c. \end{align*} $$

Combining this with (42), this shows that

$$ \begin{align*} \lim_{N\rightarrow\infty} \frac{\hat{S}_N^{k(N)}(f)(x)}{N^{\alpha} k(N)^{1-\alpha}}=c. \end{align*} $$

Finally, Lemma 19 implies that also

$$ \begin{align*} \lim_{N\rightarrow\infty} \frac{S_N^{k(N)}(f)(x)}{N^{\alpha} k(N)^{1-\alpha}}=c\quad \text{for almost every } x.\\[-41pt] \end{align*} $$

Proof. By the method of moments, it suffices to show that for all $J\geq 1$ , $0=t_0< t_1<\cdots <t_J$ and $(m_1,\ldots ,m_J) \in \mathbb {N}^J$ ,

(48) $$ \begin{align} \lim_{N \to \infty} \mathbb{E}\bigg[ \prod_{j=1}^J \binom{S_{N,j} - S_{N,j-1}}{m_j} \bigg] = \prod_{j=1}^J \frac{(t_j - t_{j-1})^{m_j}}{m_j!}, \end{align} $$

where $S_{N,j} =S_N(\chi _{N,t_j})$ .

For $N \geq \max _j m_j$ , we express the product on the left-hand side of (48) as

(49) $$ \begin{align} \prod_{j=1}^J \binom{S_{N,j} - S_{N,j-1}}{m_j} = \sum_{\mathbf{k}} \prod_{j=1}^J \prod_{i=1}^{m_j} \xi_N^{j,i}, \end{align} $$

where, with the convention $t_0=0$ and $\chi _{N,t_0}=0$ ,

$$ \begin{align*} \xi_N^{j,i}=(\chi_{N,t_j}-\chi_{N,t_{j-1}})\circ T^{k_i^j}={\mathbf{1}}_{ B_{r_N(t_j)}(x^*)\setminus B_{r_N(t_{j-1})}(x^*)}\circ T^{k_i^j}. \end{align*} $$

The sum runs over all tuples $\mathbf {k} = (k_1^1,\ldots , k_{m_1}^1,\ldots , k_1^J,\ldots , k_{m_J}^J)$ whose entries satisfy $0 \leq k_1^j < \cdots < k_{m_j}^j \leq N-1$ for each $j=1,\ldots ,J$ . Note that $\xi _N^{j,i}\cdot \xi _N^{j',i'}=0$ whenever $j\not =j'$ and $k_i^j=k_{i'}^{j'},$ and therefore, we can restrict the sum to tuples with

$$ \begin{align*} k^j_i \neq k^{j'}_{i'} \quad\text{ whenever }\; (j,i)\neq (j',i'). \end{align*} $$

For a large constant $C,$ let

$$ \begin{align*} &\Delta_N=\Delta_{N,(t_j),(m_j)}\\& \quad :=\bigg\{\mathbf{k}=(k_i^j)_{\substack{j=1,\ldots,J\\i=1,\ldots,m_j}}\,\bigg|\, \begin{aligned} &0\leq k^j_{1} < \cdots < k^j_{m_j} \leq N-1 \text{ for } j=1,\ldots,J\\ &k^j_i \neq k^{j'}_{i'} \text{ whenever }\; (j,i)\neq (j',i') \end{aligned} \bigg\} \\ \end{align*} $$

and

$$ \begin{align*} \Delta^{\prime}_N=\Delta^{\prime}_{N,(t_j),(m_j)}:=\bigg\{\mathbf{k}\in \Delta_N \,\bigg|\,\min_{\substack{j=1,\ldots,J, \; i=1,\ldots, m_j\\ j'=1,\ldots,J, \; i'=1,\ldots, m_{j'},\; (j,i)\not=(j',i')}} |k^j_i-k^{j'}_{i'}|\leq C \log N\bigg\}. \end{align*} $$

Note that $ \prod _{j=1}^J \prod _{i=1}^{m_j} \xi _N^{j,i} = 0$ if $k^j_i = k^{j'}_{i'}$ for some $(j,i)$ and $(j',i')$ with $j \neq j'$ . We then decompose the sum in (49) into two components:

$$ \begin{align*} \prod_{j=1}^J {{S_{N,j} - S_{N,j-1}} \choose m_j} = M_N+R_N, \end{align*} $$

where

(50) $$ \begin{align} M_N=M_{N,(t_j),(m_j)}= \sum_{\mathbf{k} \in \Delta_N \setminus \Delta_N'} \prod_{j=1}^J \prod_{i=1}^{m_j} \xi_N^{j,i}, \quad R_N=R_{N,(t_j),(m_j)}=\sum_{\mathbf{k} \in \Delta_N'} \prod_{j=1}^J \prod_{i=1}^{m_j} \xi_N^{j,i}. \end{align} $$

We aim to prove that

(51) $$ \begin{align} \mathbb{E}M_N \rightarrow \prod_{j=1}^{J} \frac{(t_j-t_{j-1})^{m_j}}{m_j!} \quad\text{and}\quad \mathbb{E} R_N \rightarrow 0 \quad\text{as } N \rightarrow\infty. \end{align} $$

Suppose $\mathcal {M}=\sum _{j=1}^J m_j.$ Let us first establish the asymptotic behaviour of $M_N$ . For ${\mathbf {k} \in \Delta _N \setminus \Delta _N'}$ , the indices are well separated. Analogously to Lemma 20, exponential mixing of all orders yields

$$ \begin{align*}& \bigg|\mu\bigg(\bigcap_{j=1}^J \bigcap_{i=1}^{m_j} T^{-k_i^j} (B_{r_N(t_j)}(x^*) \setminus B_{r_N(t_{j-1})}(x^*))\bigg)-\prod_{j=1}^J \mu \bigg(\!B_{r_N(t_j)}(x^*) \setminus B_{r_N(t_{j-1})}(x^*)\!\bigg)^{m_j}\bigg|\\&\quad =o(N^{-M}). \end{align*} $$

Summing over $\mathbf {k}\in \Delta _N\setminus \Delta ^{\prime }_N$ , and noting that

$$ \begin{align*} \mu (B_{r_N(t_j)}(x^*) \setminus B_{r_N(t_{j-1})}(x^*)) \sim \frac{t_j-t_{j-1}}{N} \end{align*} $$

and

$$ \begin{align*}\#(\Delta_N\setminus \Delta^{\prime}_N) \sim \prod_{j=1}^J {N \choose {m_j}},\end{align*} $$

we obtain

(52) $$ \begin{align} \mathbb{E}M_N \sim N^{-\sum_{j=1}^J m_j} \sum_{\mathbf{k}\in \Delta_N\setminus \Delta^{\prime}_N} \prod_{j=1}^J (t_j-t_{j-1})^{m_j} \sim \prod_{j=1}^{J} \frac{(t_j-t_{j-1})^{m_j}}{m_j!}. \end{align} $$

Now, we turn our attention to $R_N$ . Let $A_N = B_{r_N(t_J)}(x^*)$ . Note that $R_N$ is supported on the set

$$ \begin{align*} U_N = \bigcup_{n=0}^{N-1} T^{-n} (A_N \cap \{ \varphi_{A_N} \leq C \log N\}), \end{align*} $$

where $\varphi _{A_N}(x) = \min \{ k \geq 1 : T^k(x) \in A_N \}$ is the first return time. By the slow recurrence condition at $x^*$ ,

$$ \begin{align*} \mu(U_N) = o((\log N)^{-2\mathcal{M}}). \end{align*} $$

Applying the Cauchy–Schwarz inequality,

$$ \begin{align*} \mathbb{E} R_N\leq \mu(\mathrm{supp} (R_N))^{1/2} \| R_N \|_{L^2} \leq \mu(U_N)^{1/2} \| R_N \|_{L^2}. \end{align*} $$

Thus, to prove $\mathbb {E} R_N \rightarrow 0$ , it suffices to show $\|R_N\|_{L^2} \ll (\log N)^{\mathcal {M}}$ . We bound $\|R_N\|_{L^2}$ by noting that

$$ \begin{align*}R_N^2 \leq\sum_{0\leq k_1,\ldots,k_{2\mathcal{M}} \leq N-1} \prod_{i=1}^{2\mathcal{M}} {\mathbf{1}}_{A_N} \circ T^{k_i}= S_{N,J}^{2\mathcal{M}},\end{align*} $$

where $S_{N,J} = S_N({\mathbf {1}}_{A_N})$ and

where and are the Stirling numbers of the second kind.

Fix $m \geq 1.$ We adapt the previous notation for $t_J,m$ instead of $(t_j),(m_j).$ Then, $\Delta _N$ denotes the set of m-tuples $0 \leq k_1 < \cdots < k_m \leq N-1$ and $\Delta _N'$ is the subset of $\Delta _N$ containing tuples where the minimum distance between any two distinct indices is at most $C \log N.$ Analogously to (50), we decompose $\binom {S_{N,J}}{m} = M_{N,m} + R_{N,m}$ , where $M_{N,m}$ and $R_{N,m}$ are sums of products of ${\mathbf {1}}_{A_N} \circ T^{k_i}$ over index tuples $\mathbf {k} \in \Delta _N \setminus \Delta _N'$ and $\mathbf {k} \in \Delta _N'$ , respectively.

As in (52), $\mathbb {E}M_{N,m}$ is bounded for each fixed m. It remains to show that

$$ \begin{align*}\mathbb{E}R_{N,m}\ll (\log N)^{2\mathcal{M}}\end{align*} $$

for all $m = 1, \ldots , 2\mathcal {M}$ .

We partition $\Delta _N$ by defining

$$ \begin{align*} s(\mathbf{k}) & = \max\Big\{ s \geq 1 \mid \text{there exists }\, 1 \leq i_1 < \cdots < i_s \leq m \text{ such that }\\ & \quad \min_{r=1,\ldots,s-1} |k_{i_{r+1}} - k_{i_r}|> C \log N \Big\}. \end{align*} $$

Then, $\Delta _N = \bigcup _{s=1}^m \Delta _N^s$ , where $\Delta _N^s = \{ \mathbf {k} : s(\mathbf {k}) = s \}$ and $\Delta _N' = \bigcup _{s=1}^{m-1} \Delta _N^s$ .

For $s = 1,\ldots , m-1$ and $0\leq l_1 < \cdots < l_s \leq N-1$ with $\min _r |l_{r+1}-l_r|> C \log N$ , define

$$ \begin{align*} A_{l_1,\ldots,l_s} = \bigcap_{r=1}^s T^{-l_r} A_N. \end{align*} $$

By exponential mixing of all orders, $\mu (A_{l_1,\ldots ,l_s})\ll \mu (A_N)^s+o(N^{-s}) \ll N^{-s}$ . For ${x\in A_{l_1,\ldots ,l_s},}$ consider

$$ \begin{align*} \mathfrak{K}_{l_1,\ldots,l_s}(x)& = \left\{\mathbf{k}=(k_1,\ldots,k_m) \in \Delta_N^s\; \right.\\ & \qquad \left.\left| \begin{aligned} &\text{there exists }\, i_1,\ldots,i_s\; \text{ such that } k_{i_r}=l_r, \; \text{ for all } r=1,\ldots,s \\ &\text{and }\; x\in \bigcap_{i=1}^m T^{-k_i} A_N \end{aligned} \right.\right\}. \end{align*} $$

Then, $|\mathfrak {K}_{l_1,\ldots ,l_s}(x)| \ll (\log N)^m$ , since the remaining indices must lie within $C \log N$ of some $l_r$ . Furthermore, for any $\mathbf {k} \in \mathfrak {K}_{l_1,\ldots ,l_s}(x),$ the support of the corresponding product satisfies

$$ \begin{align*} \mathrm{supp}\bigg(\prod_{i=1}^{m} {\mathbf{1}}_{A_N}\circ T^{k_i}\bigg) \subset A_{l_1,\ldots,l_s}. \end{align*} $$

Thus,

$$ \begin{align*} \sum_{\mathbf{k} \in \Delta_N^s} \mathbb{E} \bigg[ \prod_{i=1}^m {\mathbf{1}}_{A_N} \circ T^{k_i} \bigg] & \leq \sum_{\substack{0\leq l_1<\ldots<l_s\leq N-1\\ \min_{r=1,\ldots,s-1} |l_{r+1}-l_r|>C\log N}} \int_{A_{l_1,\ldots,l_s}} |\mathfrak{K}_{l_1,\ldots,l_s}(x)| \, {d}\mu(x)\\ & \ll (\log N)^m \sum_{\substack{0\leq l_1<\ldots<l_s\leq N-1\\ \min_{r=1,\ldots,s-1} |l_{r+1}-l_r|>C\log N}} \mu(A_{l_1,\ldots,l_s})\\ &\ll (\log N)^m. \end{align*} $$

Summing over $s = 1,\ldots , m-1$ yields $\mathbb {E} R_{N,m} \ll (\log N)^m \leq (\log N)^{2\mathcal {M}},$ thereby completing the proof.

Proof. For $K=0$ , the statement reduces to Proposition 33. For $K\ge 1$ , the identities

$$ \begin{align*} \hat{S}_N^K(\chi_{N,t}) = \max(S_N(\chi_{N,t}) - K, 0) \end{align*} $$

and

$$ \begin{align*} \#(\Lambda^K \cap [0,t)) = \max(\#(\Lambda \cap [0,t)) - K, 0) \end{align*} $$

show that the conclusion follows immediately from Proposition 33.

Proof. For $\alpha>1$ , we have already shown in (53) that

$$ \begin{align*} \sum_{x\in \Lambda^K \cap [0,R)} \frac{1}{x^{\alpha}} \rightarrow \sum_{x\in \Lambda^K} \frac{1}{x^{\alpha}} \quad \text{as } R \rightarrow\infty, \, \text{almost surely}. \end{align*} $$

Therefore, the desired convergence in distribution holds with $c_R=0$ . The finiteness of moments, however, depends on the behaviour near the origin, which will be treated together with the case $\alpha \in (1/2,1]$ . Thus, in the main part of the proof, we focus on constructing the centring constants and establishing convergence for $\alpha \in (1/2,1]$ .

(i) To establish weak convergence, we first prove convergence for a suitable sequence of centring constants $c_R>0$ and a distribution $\tilde {Y}$ , without yet claiming non-degeneracy, such that

(54) $$ \begin{align} \Sigma_R - c_R \rightarrow \tilde{Y}\quad \text{as } R \rightarrow\infty, \text{ in } L^p \text{ for all } 1\leq p < \infty, \end{align} $$

where $\Sigma _R=\sum _{x\in \Lambda ^K \cap [1,R)} {1}/{x^{\alpha }}$ . In Step (iv), we will show that $c_R$ may be replaced by the explicit centring constants stated in the proposition.

Once (54) has been established, the conclusion of the proposition follows with

$$ \begin{align*} Y= \tilde{Y}+\sum_{x\in \Lambda^K \cap [0,1)} \frac{1}{x^{\alpha}}. \end{align*} $$

We separate the contribution from $[0,1)$ because it is independent of R. Thus, it does not affect the convergence as $R\to \infty $ , although it remains part of the limiting random variable and determines the range of p for which $Y\in L^p$ .

(ii) For the moment, take $c_R=\mathbb {E} \Sigma _R$ . For every even $k\geq 2$ , we will show

(55) $$ \begin{align} \mathbb{E}( |\Sigma_{R,R'} - c_{R,R'}|^k ) \rightarrow 0 \quad\text{as } R>R' \rightarrow\infty, \end{align} $$

where $ \Sigma _{R,R'} =\Sigma _R - \Sigma _{R'}$ and $c_{R,R'}=c_R-c_{R'}$ . This convergence will imply that the sequence $\{\Sigma _R-c_R\}$ is Cauchy in $L^k$ and, hence, converges in $L^k$ to some $\tilde {Y}$ . Convergence in $L^p$ for every $p\geq 1$ then follows immediately from Hölder’s inequality. Once convergence has been established, the limit $\tilde {Y}$ does not depend on p and (54) follows.

Let $\Sigma ^*_{R,R'} = \sum _{x\in \Lambda \cap [R',R)} x^{-\alpha }$ for $R>R'>1$ be the untrimmed sum and $c^*_{R,R'}=\mathbb {E} \Sigma ^*_{R,R'}$ . If $K=0$ , then $\Sigma _{R,R'}=\Sigma ^*_{R,R'}.$ If $K\ge 1,$ we observe that

$$ \begin{align*} \mathbb{P}(\Sigma_{R,R'} \not = \Sigma^*_{R,R'}) \leq \mathbb{P}( \#\{\Lambda\cap [0,R')\} <K) \ll (R')^K e^{-R'} \end{align*} $$

and, consequently,

$$ \begin{align*} &|\|\Sigma_{R,R'}-c_{R,R'}\|_{L^k}-\|\Sigma^*_{R,R'}-c^*_{R,R'}\|_{L^k} |\le\|(\Sigma_{R,R'}-c_{R,R'})-(\Sigma^*_{R,R'}-c^*_{R,R'}) \|_{L^k}\\ & \quad \le 2\|\Sigma_{R,R'}-\Sigma^*_{R,R'}\|_{L^k}\leq 2\sup |\Sigma_{R,R'}-\Sigma^*_{R,R'}|\cdot \mathbb{P}(\Sigma_{R,R'} \not=\Sigma^*_{R,R'})^{{1}/{k}}\\ & \quad \ll K(R')^{-\alpha+({K}/{k})}e^{-({R'}/{k})}. \end{align*} $$

Therefore, it suffices to prove (55) with $\Sigma ^*_{R,R'}$ instead of $\Sigma _{R,R'}$ .

Furthermore, it is enough to verify (55) for integer values $R=n$ , $R'=n'$ , as the general case follows by standard moment comparison arguments.

(iii) Let $N= |\Lambda \cap [n',n)|$ denote the number of points in $[n',n)$ , where $N\sim \text {Poisson}(n-n')$ . Given $N=k$ , the ordered arrival times $\xi _1, \xi _2,\ldots , \xi _k$ have the same distribution as k independent random variables uniformly distributed on $[n',n)$ , ordered by magnitude. Hence, $\Sigma _{n,n'}^*=\sum _{i=1}^k \xi _i^{-\alpha }$ and

$$ \begin{align*} c^*_{n,n'}=\mathbb{E} \Sigma^*_{n,n'}=\mathbb{E} [\mathbb{E} [\Sigma^*_{n,n'}\mid N]]=\int_{n'}^{n} x^{-\alpha} \, {d} x= \begin{cases} \dfrac{n^{1-\alpha} - (n')^{1-\alpha}}{1 - \alpha}, & \alpha \neq 1,\\ \log n-\log n', & \alpha = 1. \end{cases} \end{align*} $$

The moment generating function of $\bar {Y}=\Sigma ^*_{n,n'}-c^*_{n,n'}$ is defined as

$$ \begin{align*} M_{\bar{Y}}(t) = \mathbb{E}[e^{t(\Sigma^*_{n,n'}-c^*_{n,n'})}]=e^{-t c^*_{n,n'}}\mathbb{E}[e^{t \Sigma^*_{n,n'}}]. \end{align*} $$

We first compute $\mathbb {E}[e^{t \Sigma ^*_{n,n'}}]$ using the law of total expectation:

$$ \begin{align*} \mathbb{E}[e^{t \Sigma^*_{n,n'}}] = \mathbb{E}[\mathbb{E}[e^{t \Sigma^*_{n,n'}}| N]]. \end{align*} $$

As above, we obtain

$$ \begin{align*} \mathbb{E}[e^{t \Sigma^*_{n,n'}} | N=k] = \mathbb{E}\bigg[\prod_{i=1}^k e^{t \xi_i^{-\alpha}}\bigg] = (\mathbb{E}e^{t \xi_1^{-\alpha}})^k. \end{align*} $$

Since $\xi _1\sim \text {Uniform}[n', n)$ , we have

$$ \begin{align*} M(t):=\mathbb{E}[e^{t \xi_1^{-\alpha}}]=\frac{1}{n-n'} \int_{n'}^{n} e^{t u^{-\alpha}} \, {d} u. \end{align*} $$

Thus,

$$ \begin{align*} \mathbb{E}[e^{t \Sigma_{n,n'}^*}]=\mathbb{E}[M(t)^N]. \end{align*} $$

Since $N\sim \text {Poisson}(n-n')$ , its probability generating function is $\mathbb {E}[z^N] = e^{(n-n')(z-1)}$ . Substituting $z= M(t)$ , we obtain

$$ \begin{align*} \mathbb{E}[e^{t \Sigma^*_{n,n'}}]=e^{(n-n')(M(t)-1)}. \end{align*} $$

Therefore,

$$ \begin{align*} M_{\bar{Y}}(t)= e^{-t c^*_{n,n'}} e^{(n-n')(M(t)-1)}. \end{align*} $$

Substituting $c^*_{n,n'}=\int _{n'}^{n} x^{-\alpha } \, {d} x,$ we get

$$ \begin{align*} M_{\bar{Y}}(t)=\exp\bigg(\!\int_{n'}^{n} ( e^{t u^{-\alpha}}-t u^{-\alpha}-1)\, {d} u\bigg). \end{align*} $$

It follows that the cumulant generating function is given by

$$ \begin{align*} K_{\bar{Y}}(t) =\log M_{\bar{Y}}(t)= \int_{n'}^{n} ( e^{t u^{-\alpha}}-t u^{-\alpha}-1 )\, {d} u. \end{align*} $$

The kth cumulant is $\kappa _k = K_{\bar {Y}}^{(k)}(0)$ . Noting that $K_{\bar {Y}}(0) = 0$ , we further observe that ${\mathbb {E}\bar {Y}= K_{\bar {Y}}'(0) = 0}$ , which is consistent with the definition of centring. For $k \geq 2$ , define ${g(t, u) = e^{t u^{-\alpha }}-t u^{-\alpha }-1}$ . Then,

$$ \begin{align*} \frac{\partial^k}{\partial t^k} g(t, u) = (u^{-\alpha})^k e^{t u^{-\alpha}}. \end{align*} $$

Therefore,

$$ \begin{align*} K_{\bar{Y}}^{(k)}(t) = \int_{n'}^{n} u^{-k\alpha} e^{t u^{-\alpha}} \, {d} u \quad \text{and} \quad K_{\bar{Y}}^{(k)}(0) = \int_{n'}^{n} u^{-k\alpha} \, {d} u. \end{align*} $$

This yields

$$ \begin{align*} \kappa_k = \begin{cases} 0 & \text{if } k = 1, \\ \displaystyle\frac{n^{1-k\alpha} - (n')^{1-k\alpha}}{1 - k\alpha} &\text{if } k \geq 2,\, k\alpha \neq 1, \\ \log n -\log n' &\text{if } k \geq 2,\, k\alpha = 1. \end{cases} \end{align*} $$

Higher-order moments of $(\Sigma ^*_{n,n'}-c^*_{n,n'})$ can be expressed in terms of cumulants $\kappa _n$ as follows:

$$ \begin{align*} \mathbb{E}[(\Sigma^*_{n,n'}-c^*_{n,n'})^k]= \sum_{\ell=1}^k B_{k,\ell}(\kappa_1, \kappa_2,\ldots, \kappa_{k-\ell+1}), \end{align*} $$

where $B_{k,\ell }(x_1,\ldots , x_{k-\ell +1})$ denotes the partial Bell polynomial

$$ \begin{align*} B_{k,\ell}(x_1,\ldots,x_{k-\ell+1}) =\sum_{\substack{ r_1+r_2+\cdots+r_{k-\ell+1}=\ell\\ r_1+2r_2+\cdots+(k-\ell+1)r_{k-\ell+1}=k }} \frac{k!}{r_1!\,r_2!\cdots r_{k-\ell+1}!} \prod_{j=1}^{k-\ell+1}\bigg(\frac{x_j}{j!}\bigg)^{r_j}. \end{align*} $$

Given that $ \kappa _1=0 $ , all terms with $ r_1 \geq 1 $ vanish and the summation is therefore restricted to the case $ r_1 = 0 $ . This leads to

$$ \begin{align*} \mathbb{E}[(\Sigma^*_{n,n'}-c^*_{n,n'})^k]= \sum_{\substack{r_2, r_3, \cdots \geq 0 \\ 2r_2 + 3r_3 + \cdots = k}} \frac{k!}{r_2!\,r_3! \cdots} \prod_{j \geq 2} \bigg( \frac{\kappa_j}{j!} \bigg)^{r_j}. \end{align*} $$

If k is even, the dominant term corresponds to the case where $r_2={k}/{2}$ and $ r_j = 0 $ for all $ j \neq 2 $ , which is expressed as

$$ \begin{align*} \frac{k!}{({k}/{2})!\,2^{{k}/{2}}} \kappa_2^{{k}/{2}}\ll ((n')^{1-2\alpha}-n^{1-2\alpha})^{{k}/{2}}\to 0 \end{align*} $$

as $n>n'\to \infty $ since $\alpha>\tfrac 12.$

(iv) We now identify the centring constants explicitly. In the case $\tfrac 12<\alpha \le 1$ , recall that the centring term used above is $c_R=\mathbb E\Sigma _R$ . Since $\Lambda ^K$ is obtained from $\Lambda $ by removing its first K points $\xi _1,\ldots ,\xi _K$ , we have

$$ \begin{align*}\mathbb E\Sigma_R^*-c_R=\mathbb E\bigg(\sum_{j=1}^K\xi_j^{-\alpha}\mathbf 1_{{1\le\xi_j<R}}\bigg) \to\mathbb E\bigg(\sum_{j=1}^K\xi_j^{-\alpha} \mathbf 1_{{\xi_j\ge1}}\bigg)<\infty, \end{align*} $$

where, similarly to the above, $\Sigma _R^*=\sum _{x\in \Lambda \cap [1,R)} x^{-\alpha }$ denotes the untrimmed sum. Thus, $c_R$ is, up to an asymptotically constant additive correction that can be absorbed into the limiting random variable, equal to the expectation of the untrimmed sum $\Sigma _R^*$ . Choosing this normalization, we may take

$$ \begin{align*} c_R=\mathbb E\Sigma_R^* =\int_1^R x^{-\alpha}\, {d} x = \begin{cases} \displaystyle\frac{R^{1-\alpha}-1}{1-\alpha}, & \dfrac12<\alpha<1,\\[1.2ex] \log R, & \alpha=1. \end{cases}\end{align*} $$

(v) To determine the moments of the limiting random variable, recall from the preceding construction that

$$ \begin{align*} Y=\tilde Y+\sum_{x\in\Lambda^K\cap[0,1)}x^{-\alpha}. \end{align*} $$

Since $\tilde Y$ possesses finite moments of all orders, adding $\tilde Y$ does not affect the finiteness of the pth moment. Hence, the finiteness or divergence of $\mathbb {E}(|Y|^p)$ is determined by the near-origin term

$$ \begin{align*} \sum_{x\in\Lambda^K\cap[0,1)}x^{-\alpha}. \end{align*} $$

Set $p_0 = (K+1)/\alpha$ . It remains to show that this sum lies in $L^p$ for $p<p_0$ , but not for $p=p_0$ .

Condition on $\xi _{K+1}\in [1/j,1/({j-1}))$ . By the law of total expectation,

(56) $$ \begin{align} \begin{aligned} &\mathbb{E}\bigg[\bigg(\sum_{x\in\Lambda^K\cap[0,1)} x^{-\alpha}\bigg)^p\bigg]\\ & \quad = \sum_{j=2}^{\infty}\mathbb{E}\bigg[\bigg(\sum_{x\in\Lambda^K\cap[0,1)} x^{-\alpha}\bigg)^p \,\bigg|\,\xi_{K+1}\in\bigg[\dfrac1j,\dfrac1{j-1}\bigg)\bigg] \cdot \mathbb{P}\bigg(\xi_{K+1}\in\bigg[\dfrac1j,\dfrac1{j-1}\bigg)\bigg). \end{aligned} \end{align} $$

On this event, $x{\kern-1pt}\ge{\kern-1pt} 1/j$ for all $x{\kern-1pt}\in{\kern-1pt} \Lambda ^K\cap [0,1)$ , so $\sum x^{-\alpha }{\kern-1pt}\le{\kern-1pt} N j^\alpha $ with $N{\kern-1pt}=\#(\Lambda ^K\cap [0,1))$ . Conditioned on $\xi _{K+1}$ , the points in $(\xi _{K+1},1]$ form a Poisson process of mean $1-\xi _{K+1}$ , so $N=1+R$ , where R is Poisson with mean $1-\xi _{K+1}$ . Its pth moment is bounded uniformly over $\xi _{K+1}\in [1/j,1/({j-1}))$ , yielding

(57) $$ \begin{align} \mathbb{E}\bigg[\bigg(\sum_{x\in\Lambda^K\cap[0,1)}x^{-\alpha}\bigg)^p \,\bigg|\,\xi_{K+1}\in\bigg[\dfrac1j,\dfrac1{j-1}\bigg)\bigg] \ll j^{\alpha p}. \end{align} $$

For the probability factor, $\xi _{K+1}\in [1/j,1/({j-1}))$ means that there are at least $K+1$ points in $[0,{1}/({j-1}))$ , but at most K points in $[0,{1}/{j})$ . Given $\#(\Lambda \cap [0,{1}/{j}))=i$ , one needs at least $K-i+1$ points in $[{1}/{j},{1}/({j-1}))$ , with corresponding probabilities

$$ \begin{align*} \mathbb{P}\bigg(\#\bigg(\Lambda\cap\bigg[0,\frac{1}{j}\bigg)\bigg)=i\bigg)&\asymp j^{-i},\\ \mathbb{P}\bigg(\#\bigg(\Lambda\cap\bigg[\frac{1}{j},\frac{1}{j-1}\bigg)\bigg)\geq K-i+1\bigg) &\asymp j^{-2(K-i+1)}. \end{align*} $$

Summing over $i=0,\ldots ,K$ gives

(58) $$ \begin{align} \mathbb{P}\bigg(\xi_{K+1}\in\bigg[\frac{1}{j},\frac{1}{j-1}\bigg)\bigg)\asymp j^{-K-2}. \end{align} $$

Combined with (57), the jth term of the sum in (56) is $O(j^{\alpha p-K-2})$ , so the series converges precisely when $p<p_0$ .

For the lower bound of the $p_0$ th moment: if $\xi _{K+1}\in [1/j,1/({j-1}))$ , then we trivially estimate $\sum x^{-\alpha } \ge \xi _{K+1}^{-\alpha } \gg j^{\alpha }$ ; hence,

$$ \begin{align*} \mathbb{E}\bigg[\bigg(\sum_{x\in\Lambda^K\cap[0,1)}x^{-\alpha}\bigg)^{p_0} \bigg|\,\xi_{K+1}\in\left[\dfrac1j,\tfrac1{j-1}\right)\bigg] \gg j^{\alpha p_0}. \end{align*} $$

Combining this estimate with (56) and (58) yields

$$ \begin{align*} \mathbb{E}\bigg[\bigg(\sum_{x\in\Lambda^K\cap[0,1)}x^{-\alpha}\bigg)^{p_0}\bigg] \gg\sum_{j=2}^\infty j^{\alpha p_0-K-2}=\infty, \end{align*} $$

which shows the sum fails to be in $L^{p_0}$ .

Proof. For $K=0$ , we have $\hat {S}_N^0(f)=S_N^0(f)=S_N(f)$ and the claim is trivially satisfied. For the rest of the proof, we focus on the case $K\geq 1$ .

With the notation from §3.1, each $y_i$ satisfies $d(x^*,y_i)>d(x^*,\hat {x}_K)$ , since otherwise, $y_i$ would be included among the K closest points. Consequently,

$$ \begin{align*} f(y_i)&\leq (1+\epsilon(d(x^*,y_i)))\mathrm{Res}_{x^*}(f) d(x^*,y_i)^{-\beta}\\ &<(1+\epsilon(d(x^*,x_K))) \mathrm{Res}_{x^*}(f) d(x^*,\hat{x}_K)^{-\beta}. \end{align*} $$

Similarly, for each $\hat {y}_i$ , we have $d(x^*, \hat {y}_i) \leq d(x^*, \hat {x}_K)$ and, thus,

$$ \begin{align*} f(\hat{y}_i) \geq (1-\epsilon(d(x^*, \hat{x}_K)))\mathrm{Res}_{x^*}(f) d(x^*, \hat{x}_K)^{-\beta}. \end{align*} $$

The difference between the trimmed sums is

$$ \begin{align*} \hat{S}_N^K(f) - S_N^K(f) = \sum_{i=1}^m (f(y_i)-f(\hat{y}_i)). \end{align*} $$

Substituting the bounds above,

(59) $$ \begin{align} \hat{S}_N^K(f)-S_N^K(f)\leq K \mathrm{Res}_{x^*}(f)(\epsilon(d(x^*, x_K))+\epsilon(d(x^*, \hat{x}_K)))d(x^*, \hat{x}_K)^{-\beta}. \end{align} $$

Fix $\eta> 0$ and define the set

$$ \begin{align*}G_N(\eta) = \{x : d(x^*,\hat{x}_K) \geq \eta N^{-1/d},\ \max(\epsilon(d(x^*,x_K)), \epsilon(d(x^*,\hat{x}_K))) \leq \eta^{2\beta}\}.\end{align*} $$

On $G_N(\eta )$ , (59) implies

$$ \begin{align*}(\hat{S}_N^K(f)-S_N^K(f))N^{-\alpha}\leq 2K\mathrm{Res}_{x^*}(f)\eta^{\beta}<\tilde\epsilon\end{align*} $$

for $\eta $ sufficiently small. It remains to show that $G_N(\eta )$ has large probability. By Lemmas 16 and 17, for fixed K,

$$ \begin{align*} d(x^*,\hat{x}_K)\to 0,\quad d(x^*,x_K)\to 0\quad \text{for almost every } x. \end{align*} $$

Since $\epsilon (r)\to 0$ as $r\to 0$ , we get

$$ \begin{align*} \mu(\max(\epsilon(d(x^*,x_K)), \epsilon(d(x^*,\hat{x}_K))) \leq \eta^{2\beta})\to 1. \end{align*} $$

For the remaining condition, observe that

$$ \begin{align*} \{d(x^*,\hat{x}_K)<\eta N^{-1/d}\}=\{S_N({\mathbf{1}}_{B_{\eta N^{-1/d}}(x^*)})\ge K\}. \end{align*} $$

Since $S_N({\mathbf {1}}_{B_{\eta N^{-1/d}}(x^*)})\Rightarrow P(B_d\rho (x^*)\eta ^d)$ by Proposition 33, where $P(t)$ denotes a Poisson random variable with mean t,

$$ \begin{align*}\lim_{N\to\infty} \mu\{d(x^*,\hat{x}_K)<\eta N^{-1/d}\}= \mathbb{P}\{P(B_d\rho(x^*)\eta^d)\ge K \} \ll \eta^{dK}. \end{align*} $$

Combining the preceding estimates, for every $\tilde \epsilon>0$ ,

$$ \begin{align*} \limsup_{N\to\infty} \mu(\hat{S}_N^K(f)-S_N^K(f)>\tilde\epsilon N^\alpha)\ll\eta^{dK}. \end{align*} $$

Letting $\eta \to 0$ proves

$$ \begin{align*} \lim_{N\to\infty} \mu(\hat{S}_N^K(f)-S_N^K(f)>\tilde\epsilon N^\alpha)=0.\\[-41pt] \end{align*} $$

Proof of Theorem 12.

(i) We show the claims of the theorem for $\hat {S}_N^K(f)$ by proving that

(61) $$ \begin{align} \frac{\hat{S}_N^K(f)-a_N}{N^{\alpha}} \Rightarrow Y \quad\text{as } N \rightarrow\infty. \end{align} $$

Given Lemma 39, the corresponding result for $S_N^K(f)$ follows with the same normalizing constants $a_N$ .

(ii) Set $r_N=(B_d \rho (x^*) N)^{-{1}/{d}}$ and let $R>2K$ , which we regard as fixed for now. For $N \geq 1$ , define

$$ \begin{align*} \rho_N=\rho_{N,R}=R^{{1}/{d}}\, r_N \end{align*} $$

and let

$$ \begin{align*} f_N=f_{N,R}= f \cdot {\mathbf{1}}_{B_{\rho_N}(x^*)}, \quad \bar{f}_N=\bar{f}_{N,R} = f \cdot (1 - {\mathbf{1}}_{B_{\rho_N}(x^*)}). \end{align*} $$

Let $\chi _N = \chi _{N,R} = {\mathbf {1}}_{B_{\rho _N}(x^*)}$ . Then,

$$ \begin{align*} \lim_{N\rightarrow \infty} \mathbb{E}(S_N(\chi_N)) = R, \end{align*} $$

and the proof of Proposition 33, specifically (48), shows that also

$$ \begin{align*} \lim_{N\rightarrow \infty} \mathrm{Var}(S_N(\chi_N))= R. \end{align*} $$

It follows that

(62) $$ \begin{align} &\limsup_{N\rightarrow \infty} \mu(\hat{S}_N^K(f) \neq \hat{S}_N^K(f_N) + S_N(\bar{f}_N)) \leq \limsup_{N\rightarrow \infty} \mu( S_N(\chi_N) \leq K-1)\nonumber\\ & \quad \leq \limsup_{N\rightarrow \infty}\mu(|S_N(\chi_N)-\mathbb{E} (S_N(\chi_N))|\geq\mathbb{E}(S_N(\chi_N))-(K-1))\nonumber\\ & \quad \leq \lim_{N\rightarrow \infty}\frac{\mathrm{Var}(S_N(\chi_N))}{(\mathbb{E} S_N(\chi_N)-(K-1))^2} \leq \frac{4}{R}. \end{align} $$

In the remainder of the proof, we will analyse $\hat {S}_N^K(f_N)$ and $S_N(\bar {f}_N)$ separately.

Later on, in Step (v), especially (70), we will see that the contribution of $S_N(\bar {f}_N)/N^\alpha $ is asymptotically deterministic, in the sense that its fluctuations around its expectation vanish. Therefore, for the main part of the proof, we focus on $\hat {S}_N^K(f_N)$ .

(iii) Let $m\geq 1$ be a positive integer, which we regard as fixed for now, and partition the ball $B_{\rho _N}(x^*)$ into annular regions

$$ \begin{align*} A_N^j = \bigg\{x : \bigg(\frac{ R j}{m}\bigg)^{{1}/{d}}\, r_N \leq d(x, x^*) < \bigg( \frac{R (j+1)}{m} \bigg)^{{1}/{d}}\, r_N \bigg\}, \quad j=0,1,\ldots,m-1, \end{align*} $$

and introduce the trimmed counting process

$$ \begin{align*} \mathcal{N}_{N,j}^K = \hat{S}_N^K ({\mathbf{1}}_{A_N^j}). \end{align*} $$

By Proposition 37, the joint convergence

(63) $$ \begin{align} (\mathcal{N}_{N,j}^K)_{j=0,\ldots,m-1} \Rightarrow \bigg( \# \bigg\{\Lambda^K \cap \bigg[\frac{Rj}{m},\frac{R(j+1)}{m}\bigg)\bigg\} \bigg)_{j=0,\ldots,m-1} \quad\text{as } N \rightarrow\infty \end{align} $$

holds. Set $c=\mathrm {Res}_{x^*}(f) B_d^{\alpha } \rho (x^*)^{\alpha }$ and define the bounds

$$ \begin{align*} \theta_{N,m,R,-}^K = c \bigg(\frac{mN}{R}\bigg)^{\alpha} \sum_{j=0}^{m-1} \frac{1}{(j+1)^{\alpha}} \mathcal{N}_{N,j}^K \end{align*} $$

and

$$ \begin{align*} \theta_{N,m,R,+}^K = c \bigg(\frac{mN}{R}\bigg)^{\alpha} \sum_{j=1}^{m-1} \frac{1}{j^{\alpha}} \mathcal{N}_{N,j}^K. \end{align*} $$

From the joint convergence (63), we derive

(64) $$ \begin{align} N^{-\alpha} \theta_{N,m,R,\pm}^K \Rightarrow \theta_{m,R,\pm}^K \quad \text{as } N \rightarrow\infty, \end{align} $$

where

$$ \begin{align*} &\theta_{m,R,-}^K = c \bigg(\frac{m}{R}\bigg)^{\alpha} \sum_{j=0}^{m-1} \frac{1}{(j+1)^{\alpha}} \# \bigg\{\Lambda^K \cap \bigg[\frac{Rj}{m},\frac{R(j+1)}{m}\bigg) \bigg\} \\ &\theta_{m,R,+}^K = c \bigg(\frac{m}{R}\bigg)^{\alpha} \sum_{j=1}^{m-1} \frac{1}{j^{\alpha}} \# \bigg\{\Lambda^K \cap \bigg[\frac{Rj}{m},\frac{R(j+1)}{m}\bigg) \bigg\}. \end{align*} $$

Since

$$ \begin{align*} (1-\epsilon(\rho_N))c\bigg(\frac{mN}{R(j+1)}\bigg)^\alpha<f(x) <(1+\epsilon(\rho_N))c\bigg(\frac{mN}{Rj}\bigg)^\alpha \end{align*} $$

for $x\in A_{N}^{j}$ and $1\leq j\leq m-1$ , where $\epsilon $ comes from the local singularity estimate (10), and

$$ \begin{align*}f(x)>(1-\epsilon(\rho_N))c\bigg(\frac{mN}{R}\bigg)^\alpha\quad\text{for}\, x\in A_{N}^{0},\end{align*} $$

we establish the lower bound

(65) $$ \begin{align} \hat{S}_N^K(f_N) \geq (1-\epsilon(\rho_N)) \theta_{N,m,R,-}^K \end{align} $$

and the probabilistic upper bound

(66) $$ \begin{align} &\limsup_{N\rightarrow\infty}\mu(\hat{S}_N^K(f_N)>(1+\epsilon(\rho_N))\theta_{N,m,R,+}^K)\nonumber\\ &\quad\leq\limsup_{N\rightarrow\infty}\mu(\{ \text{there exists }\,0\le k\le N-1: T^k(x) \in A_N^0\})\nonumber\\ & \quad\leq \lim_{N\rightarrow\infty} N \mu(A_N^0)\leq \frac{R}{m}. \end{align} $$

(iv) Using Proposition 38, there is a non-degenerate distribution Y and constants $c_R$ such that

$$ \begin{align*} \sum_{x\in \Lambda^K \cap [0,R)} x^{-\alpha}- c_R \Rightarrow Y \quad \text{as } R \rightarrow\infty. \end{align*} $$

Moreover, for later use, we explain why the discrete lower and upper sums approximate the Poisson sum. Fix a realization in the full-probability event on which $\Lambda ^K\cap [0,R)$ is finite and does not contain $0$ . If this set is empty, the desired convergence is immediate. Otherwise, write

$$ \begin{align*} \Lambda^K\cap[0,R)=\{x_1,\ldots,x_\ell\},\quad 0<x_1<\cdots<x_\ell<R. \end{align*} $$

For all sufficiently large m, the interval $[0,R/m)$ contains no points of $\Lambda ^K$ . Hence, for every $i=1,\ldots ,\ell $ , there exists a unique $j_i\in \{1,\ldots ,m-1\}$ such that

$$ \begin{align*} \frac{Rj_i}{m}\leq x_i<\frac{R(j_i+1)}{m}. \end{align*} $$

Therefore, for all sufficiently large m,

$$ \begin{align*}\frac{1}{c}\theta^K_{m,R,-}= \sum_{i=1}^{\ell}\bigg(\frac{R(j_i+1)}{m}\bigg)^{-\alpha}, \quad\frac{1}{c}\theta^K_{m,R,+}= \sum_{i=1}^{\ell}\bigg(\frac{Rj_i}{m}\bigg)^{-\alpha}. \end{align*} $$

Since the mesh size ${R}/{m}$ tends to zero, we have ${Rj_i}/{m},({R(j_i+1)}/{m})\rightarrow x_i$ for every $i=1,\ldots ,\ell $ . Consequently,

$$ \begin{align*} \frac{1}{c}\theta^K_{m,R,-} \rightarrow \sum_{i=1}^{\ell}x_i^{-\alpha}, \quad \frac{1}{c}\theta^K_{m,R,+} \rightarrow \sum_{i=1}^{\ell}x_i^{-\alpha}. \end{align*} $$

Since the exceptional set of realizations has probability zero, we obtain

(67) $$ \begin{align} \frac{1}{c}\theta^K_{m,R,\pm} \rightarrow \sum_{x\in\Lambda^K\cap[0,R)}x^{-\alpha} \,\text{almost surely as }m\to\infty. \end{align} $$

Consequently, this convergence also holds in distribution.

(v) The conclusion now follows from a diagonal argument. For $L\ge 1$ , first choose ${\tilde R_L\ge 8L}$ sufficiently large such that

$$ \begin{align*} D_{\overline{\mathbb R}}\bigg(\sum_{x\in \Lambda^K \cap [0,\tilde{R}_L)} x^{-\alpha}-c_{\tilde{R}_L}, Y\bigg) < \frac{1}{L}, \end{align*} $$

where $D_{\overline {\mathbb R}}$ is the metric defined in (60). By (67), choose $\tilde m_L\ge 2L\tilde R_L$ sufficiently large such that

$$ \begin{align*} D_{\overline{\mathbb R}} \bigg(\frac{1}{c} \theta_{\tilde{m}_L,\tilde{R}_L,\pm}^K-c_{\tilde R_L},\sum_{x\in \Lambda^K \cap [0,\tilde{R}_L)}x^{-\alpha}-c_{\tilde R_L} \bigg) < \frac{1}{L}. \end{align*} $$

Using (62), (64) and (66), we choose $\tilde {N}_L$ sufficiently large so that for all $N\ge \tilde N_L$ , we have $\epsilon (\tilde {R}_L^{{1}/{d}}\, r_{N}) c_{\tilde {R}_L}<({1}/{L}),$ where $\epsilon $ denotes the local singularity estimate (10), and

(68) $$ \begin{align} D_{\overline{\mathbb R}}\bigg(\frac1cN^{-\alpha}\theta_{N,\tilde{m}_L,\tilde{R}_L,\pm}^K-c_{\tilde R_L}, \frac1c\theta_{\tilde{m}_L,\tilde{R}_L,\pm}^K-c_{\tilde R_L}\bigg) &< \frac{1}{L},\nonumber\\ \mu\bigg(\hat{S}_N^K(f) \neq \hat{S}_N^K(f_{N,\tilde{R}_L}) + S_N(\bar{f}_{N,\tilde{R}_L})\bigg)&<\frac{1}{L},\\ \mu\bigg(\hat S_N^K(f_{N,\tilde R_L})>(1+\epsilon(\tilde R_L^{1/d}r_N))\theta_{N,\tilde m_L,\tilde R_L,+}^K\bigg) &<\frac1L.\nonumber \end{align} $$

For $N\in [\tilde {N}_L, \tilde {N}_{L+1})$ , set $R_N=\tilde {R}_L$ and $m_N=\tilde {m}_L.$ By the choice of $\tilde m_L,\tilde R_L$ and the first line of (68), the triangle inequality gives

$$ \begin{align*} D_{\overline{\mathbb R}}\bigg(\frac{1}{c} N^{-\alpha} \theta_{N,m_N,R_N,\pm}^K - c_{R_N}, Y\bigg)<\frac{3}{L}. \end{align*} $$

Hence,

$$ \begin{align*} \frac{1}{c} N^{-\alpha}\theta_{N,m_N,R_N,\pm}^K - c_{R_N}\Rightarrow Y \quad \text{as } N \rightarrow\infty. \end{align*} $$

Using Lemma 40 with

$$ \begin{align*} X_N&=(1-\epsilon(\rho_N)) \frac{1}{c} N^{-\alpha} \theta_{N,m_N,R_N,-}^K - c_{R_N},\\ Y_N&=\frac{1}{c} N^{-\alpha} \hat{S}_N^K(f_{N,R_N}) - c_{R_N},\\ Z_N&=(1+\epsilon(\rho_N)) \frac{1}{c} N^{-\alpha} \theta_{N,m_N,R_N,+}^K - c_{R_N}, \end{align*} $$

it follows from (65) and (66) that

(69) $$ \begin{align} \frac{1}{c} N^{-\alpha} \hat{S}_N^K(f_{N,R_N}) - c_{R_N} \Rightarrow Y \quad \text{as } N \rightarrow\infty. \end{align} $$

It remains to estimate the contribution of $\bar f_{N,R_N}$ . Since now we have $N \rho _N^d\rightarrow \infty $ , we can apply Proposition 25 with the regularly adapted sequence $A_N = B_{\rho _N}(x^*)$ and ${m_1=0,m_2=2}$ ,

$$ \begin{align*} \mathrm{Var}(S_N(\bar{f}_{N,R_N})) \ll N \rho_N^{-2\beta + d} \ll N^{2\alpha} R_N^{1 - 2\alpha}, \end{align*} $$

as $N\rightarrow \infty $ . By Chebyshev’s inequality, for every $\delta>0$ and $N\in [\tilde N_L,\tilde N_{L+1})$ ,

$$ \begin{align*} \mu\bigg(\bigg|\frac{S_N(\bar f_{N,R_N})-N\mathbb{E}(\bar f_{N,R_N})}{N^\alpha}\bigg|>\delta\bigg) &\le\delta^{-2}N^{-2\alpha}\mathrm{Var}(S_N(\bar f_{N,R_N}))\\ &\ll R_N^{1-2\alpha}. \end{align*} $$

Since $\alpha>\tfrac 12$ and $R_N\to \infty $ , the right-hand side tends to $0$ . Therefore,

(70) $$ \begin{align} \frac{S_N(\bar f_{N,R_N})-N\mathbb{E}(\bar f_{N,R_N})}{N^\alpha} \to0 \end{align} $$

in probability.

Finally, combining (69), (70) and the second line of (68), we obtain